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Jet001 [13]
2 years ago
10

Lester Hollar is vice president for human resources for a large manufacturing company. In recent years, he has noticed an increa

se in absenteeism that he thinks is related to the general health of the employees. Four years ago, in an attempt to improve the situation, he began a fitness program in which employees exercise during their lunch hour. To evaluate the program, he selected a random sample of eight participants and found the number of days each was absent in the 6 months before the exercise program began and in the 6 months following the exercise program. Following are the results.
Employee Before After
1 7 7
2 7 1
3 2 6
4 5 4
5 7 3
6 6 3
7 4 1
8 1 4
At the .05 significance level, can he conclude that the number of absences has declined? Estimate the p-value.
a. State the decision rule for 0.1 significance level: H0 : μd ≤ 0; H1 : μd > 0. b. Compute the value of the test statistic.
c. Estimate the p-value? d. What is your decision regarding H0?
Mathematics
1 answer:
murzikaleks [220]2 years ago
7 0

The hypothesis is given as

H0:  μd< 0

H1: μd > 0

<h3>How to solve for the hypothesis.</h3>

The null hypothesis is given as

H0:  μd< 0

<h3>The alternative hypothesis is given as </h3>

H1: μd > 0

SWe have to find the value of s

we would use this formula s = sqrt [ (Σ(di - d)^2 / (n - 1) ]

This gives us  3.454

Next we have to determine the standard error

s / √(n)

3.454/2.8284

= 1.22

Degree of freedom = 8 - 1

= 7

t = (x1 - x2) - D / S.E

=  1.025

Find t critical at 0.10

= 1.895

P-value = 0.1697

d. Given that p value is greater thatn 0.1 we fail to reject the null hypothesis.

Read more on statistics here:

brainly.com/question/4219149

#SPJ1

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8 0
3 years ago
(a) An article in a medical journal suggested that approximately 14% of such operations result in complications. Using this esti
aev [14]

This question is incomplete, the complete question is;

Surgical complications: A medical researcher wants to construct a

99.5% confidence interval for the proportion of knee replacement surgeries that result in complications.

- An article in a medical journal suggested that approximately 14% of such operations result in complications. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.08?

Answer: a sample of operations needed is 149

Step-by-step explanation:

Given that;

confidence interval = 99.5% = 0.995 so

margin of error E = 0.08

p = 14% = 0.14

now we obtain the critical value of z t the 99.5 confidence interval

∝ = 1 - confidence interval

∝ = 1 - 0.995

∝ = 0.005

∝/2 = 0.0025

obtaining the area of probability in the right tail

Area of probability to the right is = 1 - 0.0025 = 0.9975

from probability table; critical value of t =2.81

using the formula

n = p( 1 - p ) [ (z_∝/2)/E)² ]

so we substitute  

n = 0.14 ( 1 - 0.14 ) [ 2.81 / 0.08 )²  

n = (0.14 × 0.86 ) × 1233.7656

n = 148.5453 ≈ 149

Therefore , a sample of operations needed is 149

4 0
3 years ago
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