Question

A student researcher compares the heights of men and women from the student body of a certain college in order to estimate the difference in their mean heights. A random sample of 6 men had a mean height of 68.3 inches with a standard deviation of 1.68 inches. A random sample of 11 women had a mean height of 63.2 inches with a standard deviation of 1.67 inches. Determine the 95% confidence interval for the true mean difference between the mean height of the men and the mean height of the women. Assume that the population variances are equal and that the two populations are normally distributed. Step 1 of 3: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Step 2 of 3: Find the standard error of the sampling distribution to be used in constructing the confidence interval. Round your answer to two decimal places Step 3 of 3: Construct the 95% confidence interval. Round your answers to two decimal places.

Answer #1

solution:-

given that

mean x1 = 68.3 mean x2 = 63.2

standard deviation s1 = 1.68 standard deviation s2 = 1.67

sample n1 = 6 sample n2 = 11

step 1 of 3:

degrees of freedom = (n1+n2)-2

= (6+11)-2

= 17 - 2 = 15

95% confidence with degrees of freedom = 15 is

**t = 2.131**

**step 2 of 3:**

standard error formula

sqrt(s1^2/n1 + s2^2/n2)

=> sqrt((1.68^2/6) + (1.67^2/11))

**=> 0.85**

**Step 3 of 3**

confidence interval formula

(x1-x2) +/- t * standard error

(68.3 - 63.2) +/- 2.131 * 0.85

**=> 3.29 , 6.91**

**lower limit = 3.29**

**upper limit = 6.91**

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