Question

Question 15 Type numbers in the boxes. 10 points Find the p-value for the hypothesis test. A random sample of size 50 is taken. The sample has a mean of 375 and a standard deviation of 81. H: = 400 H:u< 400 The p-value for the hypothesis test is Your answer should be rounded to 4 decimal places. Question 16 Child Health and Development Studies (CHDS) has been collecting data about Type numbers in the boxes. expectant mothers in Oakland, CA since 1959. One of the measurements taken 10 points by CHDS is the weight increase (in pounds) for expectant mothers in the second trimester. In a fictitious study, suppose that CHDS finds the average weight increase in the second trimester is 14 pounds. Suppose also that, in 2015, a random sample of 42 expectant mothers have mean weight increase of 15.7 pounds in the second trimester, with a standard deviation of 6.0 pounds. A hypothesis test is done to see if there is evidence that weight increase in the second trimester is greater than 14 pounds. Find the p-value for the hypothesis test. The p-value should be rounded to 4 decimal places.

Answer 1

Question 15

H_a : $\mu$ < 400

Left tailed test.

Given,

Hypothesized mean : $\mu _{o}$ = 400

Sample mean : $\overline{}$ $\overline{x}$ = 375

Sample size: n= 50

Sample standard deviation : s=81

-25 T-Mo Test Statistic : tstat = – s/n (375 – 400) 81/50 11.4551 = -2.1824

For left tailed test:

p-Value = P(t <tstat) = P(t < -2.1824)

Degrees of freedom =n-1 = 50-1=49

For 49 degrees of freedom, P(t<-2.1824) = 0.017

p-Value = P(t <tstat) = P(t < -2.1824) = 0.017

p-value for this hypothesis test = 0.017

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Question 16

$\mu$ : Mean weight increase in the second trimester

Null hypothesis : H_o : $\mu$ = 14

To test if there is evidence that weight increase in the second trimester is greater than 14 pounds

Alternative Hypothesis : H_a : $\mu$ > 14

Right tailed test.

Given,

Number of expectant mothers in the random sample : Sample size: n = 42

Mean weight increase in the second trimester : sample mean : $\overline{x}$ = 15.7

Sample standard deviation :s = 6.0

Hypothesized mean : $\mu _{o}$ = 14

T-HO Test Statistic : tstat = = s/n (15.7 - 14) 6/42 1.7 0.9258 = 1.8362

For right tailed test:

p-Value = P(t > tstat) = P(t > 1.8362)

degrees of freedom = n-1 =42-1=41

For 41 degrees of freedom , P(t>1.8362) = 0.0368

p-Value = P(t > tstat) = P(t > 1.8362) = 0.0368

p-value for the hypothesis test : 0.0368