Respuesta :
Answer:
A)The Null hypothesis ; H0 : u ≤ 69
The Alternate hypothesis ; H1 : u ≥ 69
B) The recruiting center claim is found at 5% significance level
C) ( 69.9488, 71.4512 )
Explanation:
A) Identify the null and alternate hypothesis
since the recruiting center is trying to test if the average height of the year's recruit is > 69. hence
The Null hypothesis ; H0 : u ≤ 69
The Alternate hypothesis ; H1 : u ≥ 69
B) Determine if the recruiters find support from the given claim at the 5% significance level
As a single tailed test we will calculate the mean and standard deviation first using MS excel
mean ( X ) = 70.7
standard deviation ( s ) = 3.02
next we will calculate the test statistic using the formula below
t = [tex]\frac{X-u}{s/\sqrt{n} }[/tex] = [tex]\frac{70.7-69}{3.02/\sqrt{64} }[/tex] = 4.503
next we will determine the P-value using MS excel
t = 4.503 , n = 64
df ( degree of freedom ) = n - 1 ( for a one tailed test )
= 64 - 1 = 63
hence the p-value at 63 degree of freedom = 0.0000148 ( using MS excel )
The p - value < significance level hence Null hypothesis is rejected while Alternate hypothesis is accepted.
The recruiting center claim is valid at 5% significance level
C) using sample data to calculate a 95% confidence interval for the average
The 95% confidence interval for the mean value u = ( 69.9488, 71.4512 )
Therefore The claim made is a reasonable one
attached below is a detailed solution
