A u.s navy recruiting center knows from past experience that the height of its recruits traditionally been distributed with mean 69 inches. The recruiting center wants to test the claim that the average height of this year's recruits is greater than 69 inches. To do so recruiting personnel to take a random sample of 64 recruits from this year and recorded their heights.
a. identify the null and alternate hypothesis.
b. Do the recruiters find support form the given claim at the 5% significance level.
c. Use sample date to calculate a 95% confidence interval for the average height .conclude?
Recruit Height
1 74.5
2 74.0
3 74.6
4 69.8
5 76.0
6 72.3
7 66.0
8 70.6
9 71.9
10 71.4
11 70.6
12 73.9
13 69.3
14 75.3
15 71.5
16 65.5
17 60.5
18 71.9
19 70.7
20 70.6
21 73.4
22 72.1
23 69.3
24 74.7
25 68.5
26 70.5
27 70.0
28 69.9
29 71.7
30 73.0
31 68.8
32 75.0
33 67.5
34 71.3
35 69.5
36 65.3
37 74.8
38 70.5
39 71.5
40 67.6
41 69.1
42 72.1
43 72.8
44 68.3
45 71.8
46 67.1
47 72.3
48 70.7
49 70.4
50 69.1
51 70.8
52 71.6
53 73.6
54 64.8
55 68.5
56 68.5
57 74.3
58 66.5
59 74.8
60 74.1
61 71.6
62 66.3
63 67.1
64 71.7

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

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b. Incurred labor costs as follows:

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