10:11 AM com deviation of 3 inches. We can also assume that it is safe to say th

Question

10:11 AM com deviation of 3 inches. We can also assume that it is safe to say that height follows a normal distribution (this may be a naïve assumption, but we will assume normality for this exercise set). You are interested in whether the female students at Ohio State are different in height from the overall 1. Assume that the population average for an American female is 64.6 inches with a population standard merican population of women. To test this hypothesis, you randomly sample 100 Ohio State female students and measure their heights and conduct a Z-test. a) What is the null hypothesis and the alternative hypothesis for this test? Assume a two-tailed test is being conducted. b) You calculate the average height from your OSU sample to be 65.3 inches. Using this sample mean and the population information given above, what is the resulting Z score for this test? 6 c) Using a normal table, find the p-value for this test. Remember, you are conducting a two-tailed d) Assume that a 0.05. Given the p-value you calculated in part c), do you conclude to re Round your answer to three decimal places. test. Round your answer to four decimal places. fail to reject the null hypothesis? What does your conclusion tell you about the heights of Ohio do you conclude to reor State female students (in this completely made-up example)? 0

Explanation / Answer

a) Let µ be the population mean height of the female students at Ohio State

Define,

Null hypothesis (H0): µ = 64.6

Alternative hypothesis (Ha): µ 64.6

b) given sample mean x = 65.3

sample size n= 100

population standard deviation = 3

Using formula of z score

z score = (x - µ) / ( / n)

              = (65.3 - 64.6) / (3 / 100)

              = 0.7 / (3 / 10)

              = 0.7 / 0.3

             = 2.333

c) for two tailed test

P-value = 2 x P( Z > | z | )

           = 2 x P (Z > 2.333)

From normal table.

P(Z < 2.33) =0.9901

=> P( Z > 2.33) = 1 - P(Z < 2.33) = 1 - 0.9901 = 0.0099

So, p-value = 2 x 0.0099 = 0.0198

• z score = 2.333

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