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nikdorinn [45]
3 years ago
8

Can you help me find the test statistic AND Pvalue for a hypothesis test. The question is: In a survey respondents were asked wo

uld you be willing to pay higher taxes if the tax revenue went directly toward deficit reduction. Treat the respondents as a simple random sample of adults.
The survey results show


FEMALE MALE


Yes 33 30

No 65 72

Total 98 102
Mathematics
1 answer:
vitfil [10]3 years ago
8 0

Answer:

Test statistic z=0.65

P-value=0.52

Fail to reject the null hypothesis.

Step-by-step explanation:

Here we have to perform a hypothesis test on the difference of proportions. We may want to answer if there is significant difference in the proportions of male and females.

The null and alternative hypothesis are:

H_0: \pi_1=\pi_2\\\\H_1: \pi_1\neq\pi_2

The significance level is defined as 0.05.

The YES proportion for females is:

p_1=33/98=0.337

The YES proportion for males is:

p_2=30/102=0.294

The weighted average of p can be calculated as:

p=\frac{n_1*p_1+n_2*p_2}{n_1+n_2}=\frac{33+30}{98+102}=0.315

With this p, we estimate the standard deviation

s=\sqrt{\frac{p(1-p)}{n_1}+\frac{p(1-p)}{n_2} }=\sqrt{\frac{0.315(1-0.315)}{98}+\frac{0.315(1-0.315)}{102} }=0.066

The test statistic z can be calculated as:

z=\frac{p_1-p_2}{s} =\frac{0.337-0.294}{0.066}=\frac{0.043}{0.066}=0.65

We can calculate the p-value for z, taking into account is a two-sided test

P(|z|>0.65)=0.52

The P-value (0.52) is greater than the significance level (0.05), so the effect is not significant. It failed to reject the null hypothesis.

We have enough evidence to conclude that the proportions that vote YES are different between genres.

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Step-by-step explanation:

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Alchen [17]

Step-by-step explanation:

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nasty-shy [4]
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5 0
3 years ago
Suppose you do not know the population mean fee charged to H&R Block customers last year. Instead, suppose you take a sample
puteri [66]

Answer:

i \to a

    n = 96040000

i \to b

    n_1 =24010000

i \to c

    n_2 =41602500

ii\toa

     E = 58.16

ii\tob

    291.84  <  \mu  < 408.16\

ii\toc

    There is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval

Step-by-step explanation:

From the question we are told that

     The sample size is n =  8

      The sample mean is  \= x  =  \$ 350    

      The sample standard deviation is  \$ 100

Considering question i

    i \to a

         At   E =  0.02  

given that the confidence level is 95%  =  0.95

         the level of significance would be  \alpha  =1-0.95 =  0.05

The critical value of  \frac{\alpha }{2} from the normal distribution table is  

        Z_{\frac{ \alpha }{2} } =  1.96

So  the sample size is mathematically evaluated as

            n = [ \frac{Z_{\frac{\alpha }{2} } *  \sigma }{E} ]^2

=>        n =[ \frac{ 1.96 *  100}{ 0.02} ]^2

=>         n = 96040000

 i \to b

  At  E_1 = 0.04    and  confidence level  = 95%  =>  \alpha_1  = 0.05   =>  Z_{\frac{\alpha_1 }{2} } =  1.96

             n_1 = [ \frac{Z_{\frac{\alpha_2 }{2} } *  \sigma }{E_1} ]^2

=>           n_1 =[ \frac{ 1.96 *  100}{ 0.04} ]^2

=>           n_1 =24010000

 i \to c

       At   E_2 =  0.04     confidence level  = 99%  =>    \alpha_2  = 0.01

The critical value of  \frac{\alpha_2 }{2} from the normal distribution table is  

        Z_{\frac{ \alpha_2 }{2} } = 2.58

=>    n_2 = [ \frac{Z_{\frac{\alpha_2 }{2} } *  \sigma }{E_2} ]^2

=>    n_2 =[ \frac{ 2.58 *  100}{ 0.04} ]^2

=>    n_2 =41602500

Considering ii

Given that the level of significance is  \alpha  = 0.10

Then the critical value  of  \frac{\alpha }{2} from the normal distribution table is  

           Z_{\frac{\alpha }{2} } =  1.645

Generally the margin of error is mathematically represented as

          E =  Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

substituting values

         E = 1.645  *  \frac{100 }{\sqrt{8} }

         E = 58.16

Generally the 90% confidence interval is mathematically evaluated as

         \= x - E <  \mu  <  \= x + E

=>      350 -  58.16  <  \mu  < 350 +  58.16

=>     291.84  <  \mu  < 408.16

So the interpretation is that there is 90% confidence that the mean  fee charged to H&R Block customers last year is in the interval .So there is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval.

8 0
3 years ago
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