Please answer all 5 questions, as I have a learning disability and is hard for m
ID: 3291279 • Letter: P
Question
Please answer all 5 questions, as I have a learning disability and is hard for me to do it. Thank you!
A union of restaurant and foodservice workers would like to estimate this year's mean hourly wage. of foodservice workers in the U.S. Last year's mean hourly wage was $8.08, and there is reason to believe that this year's value is greater than last year's The union decides to do a statistical test to see if the value has indeed increased. The union chooses a random sample of this year's wages, computes the mean of the sample to be $8.35, and computes the standard deviation of the sample to be $1.10 Based on this information, answer the questions below. What are the null hypothesis (H)and the alternative hypothesis (H) that should be used for the test? Ho: is H1 : is? 0t In the context of this test, what is a Type ll error? A Type Il error is? the hypothesis that is ? when, in fact, Suppose hat the union decides nat to ret he nul ypothesis wnat Suppose that the union decides not to reject the null hypothesis. What sort of error might it be making? ? Clear Undo HelpExplanation / Answer
(1)
Null Hypothesis Ho: µ >= 8.08
Alternative Hypothesis H1: µ < 8.08
Type II error is failing to reject the hypothesis that µ is less than $8.08 when, in fact, µ is greater than or equal to $8.08.
Suppose that the union decides not to reject the null hypothesis. What sort of error might it be making Type I.
(2)
We know that, mean µ = np
and S.D. = (npq)
Here, p = 80% = 0.8
then µ = 230*0.8 = 184
Null Hypothesis Ho: p = 0.8 i.e., Ho: µ = 184
Alternative Hypothesis H1: p 0. 8 i.e., H1: µ 184
Type of test statistic = Z
Where, Z = (np1 – np)/(np1 q1)
Here, p = 80% = 0.8
p1 = 200/230 = 0.87
n = 230
q1 = 1- p1
Value of test statistic = 3.157
Two critical values = -1.645 and 1.645
No, we cannot conclude that the proportion of high school seniors who believe that getting rich is an important goal has changed.
(3)
The null hypothesis is: H0: µ >= 505
The alternative hypothesis is: H1: µ < 505
Type of test statistic = Z
Where, Z = (X’ - µ)/(/n)
Here, X’ = 518, µ=505, = 47 and n = 100
Value of test statistic = 2.766
The p-value = 0.0028
Critical value = -2.3263
Yes, we can support the preparation course’s claim that its graduates score higher in SAT.
(4)
The null hypothesis is: H0: = 0.45
The alternative hypothesis is: H1: 0.45
Type of test statistic = t
Where, t = (X - µ)/(/n)
Here, X = 2, µ=2.2, = 0.75 and n = 10
Degree of freedom= n-1 = 9
Value of test statistic = -0.8433
Two critical values = -3.2498 and 3.2498
Yes, we can support the claim that the standard deviation of combined total of auto accidents per day differs from 0.45.
(5)
The null hypothesis is: H0: µ = 138
The alternative hypothesis is: H1: µ 138
Type of test statistic = t
Where, t = (X’ - µ)/(/n)
Here, X’ = 131, µ=138, = 15 and n = 18
Degree of freedom = n – 1 = 18-1 = 17
Value of test statistic = -1.9799
Two critical values = -1.7396 and 1.7396
No, we cannot conclude that the population mean adult sodium level differs from that claimed by the laboratory.
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