Chapter 10 hypothesis testing

Confidence Intervals for Differences

Yes. Because of the F-test and the p-value is less than 0.05 (p-value = 0.0009)

Yes, there is some cause for concern. The F-test is rather robust, however, is this case, the sample sizes are rather small and of different sizes.

These confidence intervals show that the accounting majors stating salaries, on average, are larger than the marketing majors. There is not a significant difference for the other pairs using a 95% confidence interval.

Accounting	Marketing	Finance	Management
$37,220	$28,620	$29,870	$28,600
$30,950	$27,750	$31,700	$27,450
$32,630	$27,650	$31,740	$26,410
$31,350	$27,640	$32,750	$27,340
$29,410	$28,340	$30,550	$27,300
$37,330		$29,250
$35,700		$28,890
		$30,150

45. Is there any reason to doubt the equal-variance assumption made in the one- way ANOVA model in this particular case? Explain.

There certainly is reason to doubt equal variances. The ratio of the largest standard deviation to the smallest is about 7.12, so the ratio of corresponding variances is about 51.

Source of variation	SS	df	MS	F	p-value
Between groups	140927283.143	3	46975761.048	12.677	0.0001
Within groups	77820292.857	21	3705728.231
Total variation	218747576.000	24

The ANOVA table indicates definite mean difference, even at the 1% level (since the p-value is less than .01). Even if the test is not perfectly valid (because of unequal variances), we can still be pretty confident that the means are not all equal.

The a
Accounting mean is significantly different (larger) than each of the others. Also, the Finance mean is significantly different (larger) than the Management mean. The other means are not significantly different from each other.

48. Given the information above, what is and for this comparison? Also, does this represent a one-tailed or a two-tailed test? Explain your answer.

49. What are the degrees of freedom for the t-statistic in this calculation? Explain how you would calculate the degrees of freedom in this case.

d.f = 25 + 22 – 2 = 45

The assumption is that the populations’ standard deviations are equal ().

No. There is not a statistical difference between women and men spending at Q-Mart, since p-value = 0.501 > 0.10.

No. There is not a statistical difference between women and men spending at Q-Mart, since p-value = 0.501 > 0.01.

Test statistic: Z =-1.279

There is not enough evidence to support this claim. The P-value is large (0.10).

54. Are all three sandwich shops serving the same number of customers, on average, for lunch each weekday? Explain how you arrived at your answer.

No. You should reject H_o at a 5% significance level (p-value = 0.0022). Means are not all equal.

The weights for the pooled variance are the same for each of the samples, because sample sizes are equal (sample of 10 customers from each sandwich shop).

These intervals show that there is not a significant difference between Joe’s and Bob’s. However, there is a significant difference between Joe’s and Ted’s using a 95% confidence interval.

Employee	Score before	Score after
1	62	77
2	63	77
3	74	83
4	64	88
5	84	80
6	81	80
7	54	83
8	61	88
9	81	80
10	86	88
11	75	93
12	71	78
13	86	82
14	74	84
15	65	86
16	90	89
17	72	81
18	71	90
19	85	86
20	66	92

57. Using a 10% level of significance, do the given sample data support that the firm’s training programs is effective in increasing the new employee’s working knowledge of computing?

Test statistic: t = - 4.471 (paired t-test)

The test scores have improved by an average of 11 points. Since the P-value is virtually 0, there is enough evidence to conclude that the given sample data support that the firm’s training program is increasing the new employee’s knowledge of computing at the 10% significance level.

Again, since the P-value is virtually zero, there is plenty of evidence to support the effectiveness of the program at the 1% level of significance.

Test of 1500 versus one-tailed alternative
Hypothesized mean	1500.0
Sample mean	1509.5
Std error of mean	4.854
Degrees of freedom	24
t-test statistic	1.953
p-value	0.031

59. Can the sample size be determined from the information above? Yes or no? If yes, what is the sample size in this case?

Yes. 24 + 1 = 25.

60. The firm believes that the mean life is actually greater than 1500 hours, should you conduct a one-tailed or a two-tailed hypothesis test? Explain your answer.

One-tailed, since the firm is interested in finding whether the mean is actually greater than 1500.

1509.5 hours. Yes, you would reject the null hypothesis in favor of the mean being greater than 1500 hours (0.031 < 0.05).

No. You cannot reject the null hypothesis at a 1% level of significance (0.031 > 0.01).

Test statistic: t = 1.563

The sample mean is not significantly different from 131 at even the 10% level because the p-value is greater than 0.10

64. For which values of the sample mean (i.e., average weekly grocery bill) would you decide to reject the null hypothesis at the significance level? For which values of the sample mean would you decide to reject the null hypothesis at the significance level?
ANSWER:

For either p-value (0.01 or 0.10), we find the t-value that would lead to the rejection of the null hypothesis, and then solve the equation on either side of 131. This leads to the following results:

-value	t-value	Lower limit	Upper limit
0.01	2.680	126.758	135.242
0.10	1.677	128.346	133.654

For example, at the 10% level, if we would reject the null hypothesis.

Summary statistics for two samples
	IS Salary		Mktg Salary
Sample sizes	20		20
Sample means	30401.35		27715.85
Sample standard deviations	1937.52		2983.39
Test of difference 0
Sample mean difference		2685.5
Pooled standard deviation		2515.41		NA
Std error of difference		795.44		795.44
Degrees of freedom		38		33
t-test statistic		3.376		3.376
p-value		0.0009		0.0009
Test of equality of variances
Ratio of sample variances		2.371
p-value		0.034

(2983.39)² / (1937.52)² = 2.371. Since the p-value is 0.034, you can conclude that there is a significant difference between the sample variance. They are not equal.

Conduct the t-test with individual sample variances (do not use pooled variance).

Yes. The average starting salary for IS majors is significantly larger than the starting salary for MKT majors, since p-value = 0.0009 < 0.05.

Yes. The average starting salary for IS majors is significantly larger than the starting salary for MKT majors even at a 1% significance level, since p-value = 0.0009 < 0.01.