Significant does not mean important. Never forget that even
small effects can
be statistically significant if the samples are large. To
illustrate this fact, consider
a sample of 148 small businesses. During a three-year period, 15 of
the 106
headed by men and 7 of the 42 headed by women failed.
22
(a) Find the proportions of failures for businesses headed by
women and
businesses headed by men. These sample proportions are quite close
to
each other. Give the P-value for the z test of the hypothesis that
the same
proportion of women’s and men’s businesses fail. (Use the two-sided
alternative.) The test is very far from being significant.
(b) Now suppose that the same sample proportions came from a sample
30
times as large. That is, 210 out of 1260 businesses headed by women
and
450 out of 3180 businesses headed by men fail. Verify that
the
proportions of failures are exactly the same as in part (a). Repeat
the z test
for the new data, and show that it is now significant at the α =
0.05 level.
(c) It is wise to use a confidence interval to estimate the size of
an effect
rather than just giving a P-value. Give the large sample 95%
confidence
intervals for the difference between the proportions of women’s and
men’s
businesses that fail for the settings of both parts (a) and (b).
What is the
effect of larger samples on the confidence interval?
(a)
Define the hypothesis as:
where and represents the proportion of failed business headed by men and women respectively.
It has been given that out of the businesses held by men failed, thus, the sample proportion, , is:
Also, out of the businesses held by women failed, thus, the sample proportion, , is:
The test statistic, , is computed using the formula:
where, represents the pooled proportion and is calculated using the formula:
Substitute the appropriate values in the formula to compute the test statistic:
The p-value is computed using the formula:
The value of is computed using the formula "=NORMSDIST()" in MS-excel as shown here:
Therefore,
This implies:
Since, this is a two tailed test, therefore the p-value is multiplied by 2, that is:
It is known that .
Therefore, there is an insufficient indication to discard the null hypothesis at 5% level of significance, because , that is the p-value is greater than the level of significance.
CONCLUSION:
The null hypothesis that is is true implying that the result is not significant at 5% level of significance, that is there is no difference between the proportion .
(b)
Define the hypothesis as:
where and represents the proportion of failed business headed by men and women respectively.
It has been given that out of the businesses held by men failed, thus, the sample proportion, , is:
Thus,
Also, out of the businesses held by women failed, thus, the sample proportion, , is:
Thus,
Hence, it has been verified that the proportion of failures in part(a) and part(b) are same.
The test statistic, , is computed using the formula:
where, represents the pooled proportion and is calculated using the formula:
Substitute the appropriate values in the formula to compute the test statistic:
It is known that . The critical value for the left tail of the normal distribution is fornd using the command "=NORMSINV()" in MS-EXCEL as shown below:
This implies,
Therefore, there is a sufficient indication to discard the null hypothesis at 5% level of significance, because , that is the value of the test statistic lies outside the acceptance region.
CONCLUSION:
The null hypothesis that is is not accepted in part (b) implying that there is a significant difference between the two proportions at 5% level of significance.
(c)
The % cofidence interval for the difference of proportions, say , is calculated using the formula:
where, is the right tail value of the normal distribution at significance level
The critical value for the left tail of the normal distribution is found using the command "=NORMSINV()" in MS-EXCEL as shown below:
This implies,
Since the normal distribution is symmetric, therefore, .
The 95% confidence interval for part (a) is calculated below:
The 95% confidence interval for difference in proportion, say , for part (b) is calculated below:
Therefore, the required 95% confidence intervals for part (a) and part (b) is and respectively.
It can be concluded that as a result of increasing sample size in part (b) as compared to part (a), the width of the confidence interval has been reduced due to overall fall in the value of the standard deviation of the difference in the proportions.
Significant does not mean important. Never forget that even small effects can be statistically significant if...
Never forget that even small effects can be statistically significant if the samples are large. To illustrate this fact, consider a sample of 136 small businesses. During a three-year period, 14 of the 101 headed by men and 6 of the 35 headed by women failed. (a) Find the proportions of failures for businesses headed by women and businesses headed by men. These sample proportions are quite close to each other. Give the P-value for the test of the hypothesis...
1a.) Do employees perform better at work with music playing? The music was turned on during the working hours of a business with 45 employees. Their productivity level averaged 5.2. On a different day the music was turned off and there were 40 workers. The workers' productivity level averaged 4.8. What can we conclude at the .01 level, if computed p-value is 0.12? a. Yes. Employees perform better at work when music playing. b. No. Employees don’t perform better at...
Since an instant replay system for tennis was introduced at a major tournament, men challenged 1413 referee calls, with the result that 416 of the calls were overturned. Women challenged 776 referee calls, and 225 of the calls were overturned. Use a 0.01 significance level to test the claim that men and women have equal success in challenging calls. Complete parts (a) through (c) below. a. Test the claim using a hypothesis test. Consider the first sample to be the...
1.(10) Assume that the proportion of successes in a population is p. If simple random samples of size n are drawn from the population and the proportions, p. of successes in the samples are calculated, then the distribution of the sample proportions p is normal. What are the mean and standard deviation of this Normal distribution? Hp = 2.(10) How large do the number of successes and the number of failures in a sample have to be in order to...
8. A random sample of 25 college males was obtained and each was asked to report their actual height and what they wished as their ideal height. A 95% confidence interval for μd= average difference between their ideal and actual heights was 0.8" to 2.2". Based on this interval, which one of the null hypotheses below (versus a two-sided alternative)can be rejected? A. H0: μd= 0.5 B. H0: μd= 1.0 C. H0: μd= 1.5 D. H0: μd= 2.0 9. The...
A. B. 2. The test statistic,is ______ (Round to two decimal places as needed.) 3. The P-value is ______ (Round to three decimal places as needed.) 4. State the conclusion for the test. ________________ the null hypothesis. There ______ sufficient evidence to support the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment. 5. Is it valid to argue that magnets might appear to be effective if the sample sizes...
photos for each question are all in a row (1 point) In the following questions, use the normal distribution to find a confidence interval for a difference in proportions pu - P2 given the relevant sample results. Give the best point estimate for p. - P2, the margin of error, and the confidence interval. Assume the results come from random samples. Give your answers to 4 decimal places. 300. Use 1. A 80% interval for pı - P2 given that...
Problem: Proportion of "Cured” Cancer Patients: How Does Canada Compare with Europe? Lung cancer remains the leading cause of cancer death for both Canadian men and women, responsible for the most potential years of life lost to cancer. Lung cancer alone accounts for 28% of all cancer deaths in Canada (32%. in Quebec). Most forms of lung cancer start insidiously and produce no apparent symptoms until they are too far advanced. Consequently, the chances of being cured of lung cancer...
MULTIPLE CHOICE Circle the correct answer. (2 points each) 1. A simple random sample of six patients over the age of 65 is being used in a blood pressure study. The standard error of the mean blood pressure of these six men was 22.8. What is the standard deviation of these six blood pressure measurements? a. 9.31 b. 50.98 c. 55.85 d. 136.8 e. None of the above c a. b. c. 2. A 95% CI for the ratio of...
A random sample of 88 women showed that the mean number of children reported was 1.58 with a standard deviation of 1.8. (interestingly, a sample of 88 men showed a mean of 1.3 children.) Complete parts (a) through (c) below EEB Click the lcon to view the ttable. Find a 95% confidence interval for the population mean number of children for women. Because the sample size is so large, you can use 1.96 for the critical value of t (which...