A researcher claims that the average wind speed in a certain city is 8 miles per

Question

A researcher claims that the average wind speed in a certain city is 8 miles per hour. A sample of 32 days has an average wind speed of 8.2 miles per hour, and a standard deviation of 0.5 miles per hour. The population standard deviation is 0.6 mile per hour. a. Compute a 99% confidence interval for the average wind speed. b. How large would be the sample needed at 99% confidence if the error of the sample mean with respect to the true population mean is only 0.20? c. At alpha = 0.05, is there enough evidence to reject the claim? d. Test the hypothesis that the population standard deviation is at most 0.6 miles per hour. Use alpha = 0.05.

Explanation / Answer

n = 32,

x ^ 8.2,

s = 0.6,

a) 99% confidence interval critical value when df = n-1 31 is 2.744

x^+-zvalue (s/sqrtn)

8.2+-2.744(0.6/sqrt32)

=(8.2-0.2908)(8.2+0.2908)

[7.909,8.408]

b) p^=x/n

n=x/p^= 8.2/0.2=41

c) z= (x^-u)/( s/sqrtn)=(8.2-8)/0.6/sqrt 32= 1.89

the area of z =1.89 is 0.9706

subtarct 1 -0.9706= 0.0294

since this is a two tailed test , the area of 0.0294 must be doubled the p value

the p value =0.0588

the decision is to not reject null hypothesis since the p value is greater than 0.05

d)

There is not enough evidence to reject the claim that the average wind speed is 8 miles per hour.

One-Sample Z Test of mu = 8 vs not = 8 The assumed standard deviation = 0.6

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