A new idea for treating advanced melanoma is to genetically engineer white blood

ID: 3183178 • Letter: A

Question

A new idea for treating advanced melanoma is to genetically engineer white blood cells to better recognize and destroy cancer cells, and then infuse these cells into patients. The subjects in a small initial study were 11 patients whose melanoma had not responded to existing treatments. One question was how rapidly the new cells would multiply after infusion, as measured by the doubling time in days. The data on the doubling times is found below. Use JMP to finish your calculations.

(a.) Researchers were interested specifically in whether the doubling time would be different from 1 day. What hypotheses should be used?

H0: = 1 vs. Ha: 1

H0: 1 vs. Ha: = 1

H0: X = 1 vs. Ha: X 1

H0: = 1 vs. Ha: > 1

H0: X = 1 vs. Ha: X > 1

H0: X 1 vs. Ha: X = 1

(b.) Examine the data. Is it reasonable to use the t-statistic? (Assume the 11 patients were randomly selected.)

Yes, because np and n(1-p) are both at least 10.

No, because n < 30.

No, because the distribution of doubling times is not normally distributed.

No, because np and n(1-p) are both smaller than 10.

Yes, because there are no bad outliers or skewness.

(c.) What is the average doubling time for these 11 patients? Round your answer to two decimal places.

(d.) Calculate the t-statistic for testing the above hypotheses. (You can click on the red triangle by "Doubling Times" and choose "Test Mean." Enter the hypothesized mean, and don't change anything else.) Use 4 decimal places.

(e.) What is the appropriate p-value for this test? Use 4 decimal places.

(f.) What conclusion is made?

We have evidence that the mean doubling time is equal to 1 day.

We don't have evidence that the mean doubling time is equal to 1 day.

We don't have evidence that the mean doubling time is different from 1 day.

We have evidence that the mean doubling time is different from 1 day.

(g.) Interpret the p-value above (suppose we round the actual p-value to 0.5 for simplicity).

The probability that we would observe a sample mean of 1 or farther from 1, just by chance, if the population mean were 1.08, is about 0.5.

The probability that we would observe a sample mean of 1.08 or farther from 1, just by chance, if the population mean were 1.08, is about 0.5.

The probability that we would observe a sample mean of 1 or greater, just by chance, if the population mean were 1, is about 0.5.

The probability that we would observe a sample mean of 1.08 or farther from 1, just by chance, if the population mean were 1, is about 0.5.

The probability that we would observe a sample mean of 1 or greater, just by chance, if the population mean were 1.08, is about 0.5.

Explanation / Answer

(a.) Researchers were interested specifically in whether the doubling time would be different from 1 day. What hypotheses should be used?

H0: = 1 vs. Ha: 1

H0: 1 vs. Ha: = 1

H0: X = 1 vs. Ha: X 1 , as we want to know if the value is different from 1 or not . It is non-directional test

H0: = 1 vs. Ha: > 1

H0: X = 1 vs. Ha: X > 1

H0: X 1 vs. Ha: X = 1

(b.) Examine the data. Is it reasonable to use the t-statistic? (Assume the 11 patients were randomly selected.)

Yes, because np and n(1-p) are both at least 10. ## this condition is true for binomila distribution

No, because n < 30. ## we use t stat is the sample size is less that 30 , if it is greater than 30 we use z test

No, because the distribution of doubling times is not normally distributed.

No, because np and n(1-p) are both smaller than 10.

Yes, because there are no bad outliers or skewness. ## there is no data in the question , hence we cant be sure of this

(c.) What is the average doubling time for these 11 patients? Round your answer to two decimal places.

## no data in the question to answer this

(d.) Calculate the t-statistic for testing the above hypotheses. (You can click on the red triangle by "Doubling Times" and choose "Test Mean." Enter the hypothesized mean, and don't change anything else.) Use 4 decimal places.

## no data

(e.) What is the appropriate p-value for this test? Use 4 decimal places.

## the p value can be calculated from the t stat . we first calculate the degree of freedom as n-1 , where n is the number of observation. so once we know the degree of freedom we look the p value in the t table. if the p value is less than 0.05 , then we reject the null hypothesis in favor of alternate hypothesis.

(f.) What conclusion is made?

if the p value is less than 0.05 , then we reject the null hypothesis in favor of alternate hypothesis.

if the p value is greater than 0.05 , then we accept the null hypothesis ## we cant answer this unless we have the data