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Vladimir79 [104]
2 years ago
12

Some college graduates employed full-time work more than hours per week, and some work fewer than hours per week. We suspect tha

t the mean number of hours worked per week by college graduates, , is different from hours and wish to do a statistical test. We select a random sample of college graduates employed full-time and find that the mean of the sample is hours and that the standard deviation is hours.
Mathematics
2 answers:
Alex_Xolod [135]2 years ago
5 0

Answer:

a)Null hypothesis:\mu =40      

Alternative hypothesis:\mu \neq 40      

b) A Type of error I is reject the hypothesis that \mu is equal to 40 when is fact \mu is equal to 40c) We can commit a Type II Error, since by definition "A type II error is the non-rejection of a false null hypothesis and is known as "false negative" conclusion"Step-by-step explanation:Assuming this problem: "Some college graduates employed full-time work more than 40 hours per week, and some work fewer than 40 hours per week. We suspect that the mean number of hours worked per week by college graduates, [tex]\mu , is different from 40 hours and wish to do a statistical test. We select a random sample of college graduates employed full-time and find that the mean of the sample is 43 hours and that the standard deviation is 4 hours. Based on this information, answer the questions below"

Data given

\bar X=43 represent the sample mean

\mu population mean (variable of interest)

s=4 represent the sample standard deviation

n represent the sample size  

Part a: System of hypothesis

We need to conduct a hypothesis in order to determine if actual mean is different from 40 , the system of hypothesis would be:    

Null hypothesis:\mu =40      

Alternative hypothesis:\mu \neq 40      

Part b

In th context of this tes, what is a Type I error?

A Type of error I is reject the hypothesis that \mu is equal to 40 when is fact [tex]\mu is equal to 40

Part c

Suppose that we decide not to reject the null hypothesis. What sort of error might we be making.

We can commit a Type II Error, since by definition "A type II error is the non-rejection of a false null hypothesis and is known as "false negative" conclusion"

Anika [276]2 years ago
3 0

Answer:

Step-by-step explanation:

What are the null hypothesis (H0) and the alternative hypothesis (H1) that should be used for the test?

A: H0: μ is  LESS THAN OR EQUAL TO

    H1: μ is  GREATER THAN

B: In the context of this test, what is a Type II error?

A Type II error is FAILING TO REJECT the hypothesis that μ is LESS THAN OR EQUAL TO  when, in fact, μ is GREATER THAN .

C: Suppose that we decide not to reject the null hypothesis. What sort of error might we be making? TYPE II

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Given:

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Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

The sample proportion is :  \hat p = x/n = 1219/3532 = 0.345

Firstly, the pivotal quantity for 95% confidence interval for the proportion of the population is given by;

      P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion = 0.345

           n = sample of participants = 3532

           p = population proportion

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population​ proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level of

                                                         significance are -1.96 & 1.96}

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u>95% confidence interval for p</u>= [\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

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Hence, 95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

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