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Child Health and Development Studies (CHDS) has been collecting data about expectant mothers in Oakland, CA since 1959. One of the measurements taken 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 40 expectant mothers have mean weight increase of 16 pounds in the second trimester, with a standard deviation of 6 pounds. At the 5% significance level, we can conduct a one-sided T-test to see if the mean weight increase in 2015 is greater than 14 pounds. Statistical software tells us that the p-value = 0.021.Which of the following is the most appropriate conclusion?There is a 2.1% chance that a random sample of 40 expectant mothers will have a mean weight increase of 16 pounds or greater if the mean second trimester weight gain for all expectant mothers is 14 pounds.There is a 2.1% chance that mean second trimester weight gain for all expectant mothers is 14 pounds in 2015.There is a 2.1% chance that mean second trimester weight gain for all expectant mothers is 16 pounds in 2015.There is 2.1% chance that the population of expectant mothers will have a mean weight increase of 16 pounds or greater in 2015 if the mean second trimester weight gain for all expectant mothers was 14 pounds in 1959.Find the p-value for the hypothesis test. A random sample of size 50 is taken. The sample has a mean of 420 and a standard deviation of 81.H0: µ = 400Ha: µ > 400The p-value for the hypothesis test is

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KhiradAfaq

We accert H1 alternative that is result is population mean greater than 400.

There is a 2.1% chance than mean second Trimester weight gain for all expectant mothers is 14 pounds in 2015.

What is standard deviation?

The standard deviation is a statistic that expresses how much variance or dispersion there is in a group of numbers. While a high standard deviation suggests that the values are dispersed throughout a wider range, a low standard deviation suggests that the values tend to be close to the established mean.

We have given: μ = 14, x bar = 16, n = 40, s = 6,  α = 5%, = 0.05

To test μ0 : μ = 14   Vs  H1: μ > 14

Test statistics = (X bar - μx)√n / sx = (2 x 6.3245) / 6 = 2.1081

So, P value = 0.021

Conclusion: There is a 2.1% chance than mean second Trimester weight gain for all expectant mothers is 14 pounds in 2015.

P value for the hypothesis test

n = 50, x bar = 420, sx = 81, μ = 400,

To test:  H0: μ 400    Vs   H1:  μ > 400

It is right tail test

Test statistics =  (X bar - μx)√n / sx = (420 - 400) √50 / 81 =

z = 1.7459

P value = 0.04093

Declsion: We reject H0 is p value < α

Here α = 0.05

Hence P value < α

Conclusion: We accert H1 alternative that is result is population mean greater than 400.

Hence, we accert H1 alternative that is result is population mean greater than 400.

There is a 2.1% chance than mean second Trimester weight gain for all expectant mothers is 14 pounds in 2015.

To know more about standard deviation, click on the link

https://brainly.com/question/475676

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