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General Statistics Problems

See attached file for full details of problems.

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9.69 Late payments of medical claims can add to the cost of health care. An article reported that for one insurance company, 85.1% of the claims were paid in full when first submitted. Suppose that the insurance company developed a new payment system in an effort to increase this percentage. A sample of 200 claims processed under this system revealed that 180 of the claims were paid in full when first submittted.

a. At the 0.05 level of significance, is there evidence that the proportion of claims processed under this new system is higher than the article reported for the previsou system?

b. Compute the p-value and interpret its meaning.

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9.72 A Wall Street Journal poll asked respondents if they trusted energy-efficiency ratings on cars and appliances; 552 responded yes and 531 responded no.

a. At the 0.05 level of significance, use the six-step critical value approach to hypothesis testing to try to prove that the percentage of people who trust energy-efficiency ratings differ from 50%.

b. Use the five step p-value approach.

c. Intepret the results

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9.73 One of the biggest issues facing e-retailers is the ability to reduce the proportion of customers who cancel their transactions after they have selected their products. It has been estimated that about 1/2 of prospective customers cancel their transactions after they have selected their products. Suppose a company changed its Web site so that customers could use a single-page checkout process rather than multiple pages. A sample of 500 customers who had selected their products were provided with the new checkout. Of these 500 customers, 210 canceled their tranactions after they have selected their products.

a. At the 0.01 level of significance, is there evidence that the proportion of customers who selected products and then cancelled their transaction was less than 0.50 with the new system?

b. Suppose that a sample of 100 customers who had selected their products were provided with the new checkout system and that 42 of those customers cancelled their transactions after they had selected their products. At the 0.01 level of significance, is there evidence that the proportion of customers who selected products and then cancelled their transaction was less than 0.50 with the new system?

c. Compare the results of (a) and (b) and discuss the effect that sample size has on the outcome, and, in general on hypothesis testing.

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9.74 More professional women then everbefore are forgoing motherhood because of the time constraints of their careers. However, many women still manage to find time to climb the corporate ladder and set time adie to have children. A survey of 187 attendees at Fortune Magazine's Most Powerful Women in Business summit in March 2002 found that at least 133 had at least one child. Assume that the group of 187 women was a random sample from the population of all successful women executives.

a.What was the sample proportion of successful woemn executives who had children?

b. At the 0.05 level of significance, can you state that mnore than half of all successful women execurtives had children?

c. At the 0.05 level of significance, can you state that mnore than two-thirds of all successful women executives had children?

d. Do you think the random sample assumption is valid? Explain.

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10.28 Multiple myeloma, or blood plasma cancer, is characterized by increased blood vessel formation in the bone marrow that is a prognostic factor in survival. One treatment approach used for multiple myeloma is stem cell transplantation with the patient's own stem cells. The data stored in the file myeloma.xls represent the bone marrow microvessal density for patients who had a complete response to the stem cell transplant, as measured by blood and urine tests. The measurements were taken prior to the stem cell transplat and at the time of the complete response.

a. At the 0.05 level of significance,is there evidence that the mean bone marrow microvessel density is higher before the stem cell transplant than after the stem cell transplant?

b. Interpret the meaning of the p-value in (a).

c. Construct and interpret a 95% confidence interval estimate of the mean difference in bone marrow microvessel density before and after the stem cell transplant.

d. What assumption is necessary to perform the test in (a)?

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10.29 Over the past year, the vice president for human resources at a large medical center has run a series of three month workshops aimed at increrasing worker motivation and performance. To check the effectiveness of the workshops, she selected a random sample of 35 employees from the personnel files and recorded their most recent annual performance ratings, along with their ratings prior to attending the workshops. The data are stored in the file perform.xls. The Microsoft excel results in Panels A and B in the text provide both descriptive and inferential information so that you can analysze the results and examine the assumptions of the assumptins of the hypothesis test used: State your findings in a report to the vice president for human resources.

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10.82 Problem 1.27 on page 15 describes a survey of 50 undergraduate students (see the file undergradsurvey.xls.)

a. For these data, at the 0.05 level of significance, is there evidence of a difference between males and females in grade point average, expected starting salaryt, salary expected in five years, age, and spending on terxtbooks and supplies?

b. For these data, at the 0.05 level of significance, is there evidence of a difference between those students who plan to go on to graduate school and those who do not plan to go on to graduate school in in grade point average, expected starting salary, salary expected in five years, age, and spending on terxtbooks and supplies?

Attachments

Solution Preview

The answers to all of the questions are below and in the "Support data" worksheets in the Excel file.

9.69

In this problem, we are comparing the new percentage of paid claims to the old percentage. We will use a z-test.
The old percentage: 85.10%
The new percentage (180/200 = 0.9): 90%

We will use a one-sided z-test to determine if there is evidence that the proportion of claims processed under this new system is higher than the article reported for the previous system.

The null hypothesis is that the proportion under the new system is less than or equal to the proportion under the old system. H0: p <= p0

The alternative hypothesis is that the proportion under the new system is greater than the proportion under the old system. Ha: p > p0

The z-statistic can be calculated using the following formula: [see attachment]

Using the data from this problem, our observed value of z is z = 2.3:

z = (0.90 - 0.851) / sqrt( 0.90*0.10/200 ) = 2.309882152

This can be converted into a p-value. I used the calculator here: http://faculty.vassar.edu/lowry/ch6apx.html.
The one-tailed p-value associated with z = 2.3 is p = 0.0107.

Because the observed p-value of p = 0.0107 is less than 0.05, we can reject the null hypothesis at the 0.05 level.

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9.72

Six Step Method:

Step 1: State the null and alternative hypotheses.

Null hypothesis: The percentage of people who responded yes is equal to 50%. H0: p = 0.50.
Alternative hypothesis: The percentage of people who responded yes is different than 50%. Ha: p =/ 0.50.

Step 2: Choose a level of significance.

The significance level is 0.05.

Step 3: Determine the critical values.

The critical values for a two tailed 0.05 significance level are z = 1.96 and z = -1.96.

Step 4: Calculate the test statistic.

This uses the same formula for z as in problem 9.69.

Sample proportion = 552/1083 = 0.509695291
Sample size = 552 + 531 = 1083

z = (0.509695 - 0.50) / sqrt( (0.509695)*(1-0.509695)/1083 ) = 0.638224827

Step 5: Compare the test statistic to the critical values.

The observed value of z is less than the critical value of z. Therefore, we cannot reject the null hypothesis.

Step 6: Conclusions

We cannot reject the null hypothesis. There is no evidence to suggest that the number of people who answered "yes" is different than 50%.

Five Step Method:

Step 1: State the null and alternative hypotheses.

Null hypothesis: The percentage of people who responded yes is equal to 50%. H0: p = 0.50.
Alternative hypothesis: The percentage of people who responded yes is different than 50%. Ha: p =/ 0.50.

Step 2: Choose a level of significance.

The significance level is ...

$2.19