Tests of the true value of an unknown population mean can be either one-tailed (left- tailed or right-tailed) or two-tailed.
1. two-tailed (non-directional)
H0: μ = μ0 ← a possibility you want to test (null hypothesis)
Ha: μ ≠ μ0 ← what the sample evidence suggests (alternative hypothesis)
Reject H0 if ( x – μ0) is a large positive number or a large negative number
2. left-tailed
H0: μ ≥ μ0} Reject H0 if ( x – μ0) is a large negative number. Ha: μ < μ0}
3. right-tailed
H0: μ ≤ μ0} Reject H0 if ( x – μ0) is a large positive number. Ha: μ > μ0}
Example: n=50, σ = 4.6, x =6.
1. H0:μ=5 Ha: μ ≠ 5
∝ = .05
Ζ = 6 – 5 = 1 = 1.54
.65 .65
σ / n = 4.6/ 50
±Ζα/2 =±1.96
2-tailed rejection region at .05
Because Ζ = 1.54 is not In the rejection area, fail to reject H0.
The probability of being wrong in this conclusion is called p-value:
p = P(Ζ>1.54) + P(Ζ<-1.54)
= .0618 x .0618
(for 2-tailed test) p = .1236 > .05
Ζα =1.645 Because Ζ = 3.08
is in the rejection region, reject H0.
p = P (Ζ>3.08) p = .001 < .05
2. H0:μ≥6.5 Ha: μ < 6.5
∝ = .05
Ζ = 6 – 6.5 = -.5 = -.77
.65 .65
3. H0:μ ≤ 4 Ha: μ > 4
∝ = .05
Ζ = 6 – 4 = 2 = 3.08
.65 .65
-Ζα =-1.645
Both are 1-tailed rejection region at .05
Because Ζ = -0.77 is
not in the rejection region, fail to reject H0.
p = P (Ζ<-0.77) p = .2206 > .05
Remember 1: When looking-up the proportion in the tail in the Unit Normal Z table, the given p-value is for one-tailed tests. If you have a two-tailed test, as seen in example 1 on the previous page, multiply the given p-value by 2 to reflect the two-tailed nature of the test.
Remember 2: SPSS’ p-values are presented as derived from two-tailed tests. If your alternative hypothesis being tested reflects a one-tailed test, you must divide the given SPSS p-value by 2 to reflect the one-tailed nature of your alternative hypothesis. From example 1 on the previous page, the p-value of .1236, reflecting a two-tailed test, would be readjusted by .1236 / 2 = one-tailed p-value of .0618.
Testing Hypotheses: σ is known and n > 30 I. One sample test for the population mean
1. H0: μ=μo
2. Ha: μ≠μo;μ<μo;μ>μo
3. Test statistic: Assume that H0 is true and see if you have enough data/evidence to reject it.
3B. How far sample mean x is from μ
z = x − μo
σ
n
4. P-value: Probability when H0 is true of getting a test statistic as extreme as you did. Assume that H0 is true until disproved.
5. Conclusion:
A. Small p-value = reject H0, there is enough evidence to say that Ha is true
x is so far form μo that it is highly unlikely μo is true.
B. Large p-value = fail to reject H0. There is not sufficient evidence to say Ha is
true.
1. Example:
The level of calcium in the blood of healthy, young adults varies with a mean of 9.5 mg per deciliter and a SD of 0.4. A clinic in rural Illinois measures the blood calcium level of 180 healthy pregnant women and finds x = 9.57mg. Is this an indication that the mean calcium level in this population differs from 9.5mg?
H0: μ = 9.5
Ha: μ ≠ 9.5
Z= x−μo 9.57−9.5=.07=2.33
σ n .4 180 .03
With alpha = .05 and because this is a two-tailed test (i.e., the = and the ≠ ), the critical region would consist of a Z score beyond ± 1.96 (note: this is found in the proportion of the tail, where .0250 is closest to Z = 1.96, so .0250 x 2 = .05).
Thus, with our Z value of 2.33 (look at table), the p-value is determined by .0099 (proportion of the tail section) x 2 (because of the two-tailed nature of this test) = .0198
There is only a 1.98% chance of getting a x of 9.57 or more extreme.
Conclusion:
At the .05 level, we would reject H0 and say there is enough evidence to show the mean is different from 9.5. Thus, we have shown that the average level of calcium in the blood of pregnant Illinois women is different from 9.5 (or all other healthy young adults).
II. Example:
Mike gave the SAT math test to a simple random sample of 500 seniors from Illinois. These students had a mean score of 461 ( x ). Is this good evidence that the mean for all Illinois seniors is > 450. σ = 100
H0: μ ≤ 450
Ha: μ > 450
z= x−μ=461−450= 11 =2.46
σ n 100 500 4.472
With alpha = .05 and because this is a one-tailed test (i.e., the ≤ and the >), the critical region would consist of a Z score beyond ± 1.645 (i.e., because a Z value of 1.64 and 1.65 are of equal distance in this case, we take the average).
Thus, with a Z value of 2.46, the p-value = .0069.
We reject H0 at the .05 level and say that there is a less than 1% chance of getting a
x of 450 or more extreme.
Conclusion:
A Z value of 2.46 indicates that our sample mean is in the critical region and this is a very unlikelyoutcomeifH0 istrueand,thus,thedecisiontorejectH0.Themeantestsco
Part A. Multiple Choice Questions. For each question, you are encouraged to give a reason or show work for
partial credit. You must show your work or reason if the question is marked with an asterisk (*).
1. Confidence intervals are useful when trying to estimate _______.
a. unknown parameters
b. known parameters
c. unknown statistics
d. known statistics
2. The one-sample z statistic is used instead of the one-sample t statistic when ______.
a. μ is known
b. μ is unknown
c. σ is known
d. σ is unknown
3. The ____ the P-value, the stronger the evidence against the null hypothesis provided by the data.
a. larger
b. smaller
4. (*) The test statistic for a two-sided significance test for a population mean is z = –2.12. What is the
corresponding P-value?
a. 0.017
b. 0.034
c. 0.483
d. 0.983
Answer: P(Z < -2.12) = 0.0170. Since this is a two-sided test, P-value is twice as much to include the other tail. 0.0170 * 2 = 0.034.
5. The probability you reject the null hypothesis when in fact the null hypothesis is true is called __________.
a. 4.6
b. 15.9 c. 1.3
d. 5.8
Answer: 𝑆𝑆𝑆𝑆 = 25.8 = 25.8 = 5.8 e. 2.6 𝑥𝑥̅ √20 4.472
a. a Type I error
b. a Type II error
c. the power
6. (*) A random sample of 20 observations produced a sample mean of 𝑥𝑥̅ = 92.4 and s = 25.8. What is the value of the standard error of 𝑥𝑥̅?
7. (*) The heights (in inches) of adult males in the United States are believed to be Normally distributed with mean μ. The average height of a random sample of 25 American adult males is found to be 𝑥𝑥 ̅ = 69.72 inches, and the standard deviation of the 25 heights is found to be s = 4.15. A 90% confidence interval for μ is
a. 69.72 ± 1.09
b. 69.72 ± 1.37
c. 69.72 ± 1.42
Answer: The t critical value for 90% CI for df = 24 is 1.711. So margin_or_error = (1.711) ∙ 4.15 = (1.711) ∙ (0.83) = 1.42
√25
8. Suppose we were interested in determining if there were differences in the average prices among two local supermarkets. We randomly pick six items to compare at both supermarkets. Which statistical procedure would be best to use for this study?
a. Matched-pairs t procedure if you interpreted as ‘same six items’
b. One-sample t test
c. Two-sample t test if you interpreted as ‘different six items’
d. None of the above
9. (*) Perform a one-sample t-test using the following statistics: n = 5 𝑥𝑥̅ = 3.871 s = 0.679
The null hypothesis is μ = 5.0 is
a. accepted at the 5% level; accepted at the 1% level.
b. accepted at the 5% level; rejected at the 1% level.
c. rejected at the 5% level; accepted at the 1% level.
d. rejected at the 5% level; rejected at the 1% level.
Answer: Test statistic: 𝑡𝑡 = 𝑥𝑥̅−𝜇𝜇 = 3.871−5.0 = −3.71 𝑠𝑠 0.679
√𝑛𝑛 √5
Assuming a two-sided test, for df = 4,
• 95% CI of the t-statistic assuming H0 is [-2.776, 2.776]
• 99% CI of the t-statistic assuming H0 is [-4.604, 4.604]
Therefore we reject H0 for alpha = 0.05 but not for alpha 0.01.
10. (*) You buy a package of 122 Smarties and 19 of them are red. What is a 95% confidence interval for the true proportion of red Smarties?
So the 95% CI is [0.156 – 0.0647, 0.156 + 0.0647], which yields [0.0913, 0.2207].
a. (0.092, 0.220)
b. (0.103, 0.230)
c. (0.085, 0.199)
Answer: 𝑝𝑝̂ = 19 = 0.156. margin_of_error = (1.96) ∙ �(0.156)∙(0.844) = (1.96) ∙ (0.033) = 0.0647. 122 122
11. (*) It is thought that 12% of all students taking a particular course received a grade of A. In a sample of 155 students, it is found that 21 made an A. What is the test statistic for testing the true proportion is 12%?
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Part B. Short Answer Questions
1. [Textbook Exercise 5.30, 5.32] A tablet PC contains 3217 music files. The distribution of file size is highly skewed with many small file sizes. Assume that the standard deviation for this population is 3.25 megabytes (MB).
a. 0.53
b. 0.01
c. 0.577
Answer: 𝑝𝑝̂ = 21 = 0.135. 𝑍𝑍 = 0.135−0.12 = 0.015 = 0.577 155 �(0.12)∙(0.88) 0.026
155
a. What is the standard deviation of the average file size when you take an SRS of 25 files from this
population?
Answer: 𝜎𝜎 = 325 = 0.65
𝑥𝑥̅
√25
b. Suppose you plan to take an SRS of n = 50 files instead. Explain why it may be reasonable to assume that the average 𝑥𝑥̅ is approximately Normal even though the population distribution is highly skewed.
Answer: Because of the Central Limit Theorem (CLT), we can assume the sample mean is normally distributed when the sample size is large enough. In this case, n = 50 is greater than 30, thus the CLT applies.
2. [Textbook Exercise 7.57] For each of the following, answer the question and give a short explanation of your reasoning.
a. A significance test for comparing two means gave t=−1.97 with 10 degrees of freedom. Can you reject the null hypothesis that the μ’s are equal versus the two-sided alternative at the 5% significance level?
Answer:
Probability for t = -1.97 for df = 10 is between 0.025 and 0.05. So for two-sided test, the p-value is between 0.05 and 0.1. Therefore we fail to reject H0 at the significance level alpha = 0.05.
The graph on the left is calculated by a software. You see the exact P-value (0.07714).
b. Answer part (a) for the one-sided alternative that the difference between means is negative.
Answer: If the alternative hypothesis was Ha: μ1 – μ2 < 0, we can reject the null hypothesis at alpha = 0.05 level because the p-value would be between 0.025 and 0.05.
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Part C. Problems
1. One-sample t-test. To test the hypothesis that eating fish makes one smarter, a random sample of 12
persons take a fish oil supplement for one year and then are given an IQ test. Here are the results: 116 111 101 120 99 94 106 115 107 101 110 92
Test using the following hypotheses, report the test statistic with the P-value, then summarize your conclusion.
H0: μ = 100 Ha: μ > 100
Answer: Hypotheses:
H0: μ = 100 (no effect — eating fish does not help increase the mean IQ) Ha: μ > 100 (effect — eating fish helps increase the mean IQ)
Test statistic:
From the d𝒔𝒔ata, we obtain 𝒙𝒙� = 𝟏𝟏𝟏𝟏𝟏𝟏 and 𝐬𝐬𝒙𝒙 = 𝟖𝟖. 𝟖𝟖𝟖𝟖. Then we get
𝒕𝒕= �𝒙𝒙−𝝁𝝁𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟏𝟏 =𝟏𝟏.𝟖𝟖𝟓𝟓. 𝒙𝒙 𝟖𝟖.𝟖𝟖𝟖𝟖 𝟏𝟏.𝟓𝟓𝟓𝟓
P-value: √𝒏𝒏
√𝟏𝟏𝟏𝟏
Because n = 12, we use the t distribution with df = 11 to find the probability. According to Table D, for t = 2.35 for df = 11, the probability is between 0.01 and 0.02. Since this is a one-sided (upper-tail) test, P-value is between 0.01 and 0.02.
Conclusion:
Since values between 0.01 and 0.02 are < 0.05, we reject the null hypothesis at the significance level
0.05, and conclude that the fish oil supplement did make a significant increase in the mean IQ.
2. Matched one-sample t-test. The water diet requires you to drink 2 cups of water every half hour from when you get up until you go to bed but eat anything you want. Four adult volunteers agreed to test this diet. They are weighed prior to beginning the diet and 6 weeks after. Their weights in pounds are
Person Weight before Weight after Difference
1 2 180 125 170 130
3 4 240 150 215 152
mean _ s.d. 173.75 49.56 166.75 36.09
10 -5
Conduct a one-sample t-test using the difference with the following hypotheses:
H0: Diff = 0 Ha: Diff ≠ 0
Report the test statistic with the P-value, then summarize your conclusion.
Answer: Hypotheses:
H0: Diff = 0 (no difference — there is no difference in the mean of the weight difference) Ha: Diff ≠ 0 (difference – diet made difference in the mean of the weight difference)
25 -2
7 13.64
Test statistic:
From the d𝒔𝒔ata, we know 𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫 = 𝟕𝟕 and 𝐬𝐬𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫 = 𝟏𝟏𝟖𝟖. 𝟏𝟏𝟔𝟔. Then we get
�������
�������
𝒕𝒕= 𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫−𝝁𝝁𝟏𝟏 = 𝟕𝟕−𝟏𝟏 = 𝟕𝟕 =𝟏𝟏.𝟏𝟏𝟏𝟏𝟏𝟏. 𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫 𝟏𝟏𝟖𝟖. 𝟏𝟏𝟔𝟔 𝟏𝟏. 𝟖𝟖𝟏𝟏
√𝒏𝒏 √𝟔𝟔
P-value:
Because n = 4, we use the t distribution with df = 3 to find the probability.
According to Table D, for t = 1.026 for df = 3, the probability is between 0.15 and 0.20. Since this is a two-sided test, P-value is between 0.30 and 0.40.
Conclusion:
Since values between 0.30 and 0.40 are > 0.05, we fail to reject the null hypothesis at the 0.05 significance level. We do not have enough evidence to conclude that the water diet has an impact on the weight.
3. Two-sample t procedure. Two different alloys are being considered for making lead-free solder used in the wave soldering process for printed circuit boards. A crucial characteristic of solder is its melting point, which is known to follow a Normal distribution. A study was conducted using a random sample of 21 pieces of solder made from each of the two alloys. In each sample, the temperature at which each of the
21 pieces melted was determined. The mean and standard deviation of the sample for Alloy 1 were x1 = 218.9oC and s1 = 2.7oC; for Alloy 2 the results were x2 = 215.5oC and s2 = 3.6oC. If we were to test H0:
μ1 = μ2 against Ha: μ1 ≠ μ2 , what would be the value of the test statistic??
Answer:
The two-sample test statistic (for the hypothesis 𝜇𝜇1 − 𝜇𝜇2 = 0) is
𝑡𝑡=(�𝑥𝑥��−�𝑥𝑥��)= 218.9−215.5 = 3.4 = 3.4 = 3.4 =3.46
1 2 (2.7)2 (3.6)2 √0.347 + 0.617 √0.964 0.982
12
�𝑠𝑠2+𝑠𝑠2 � 21 + 21
𝑛𝑛𝑛𝑛 12
And the degrees of freedom is 21 – 1 = 20.