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9966 [12]
3 years ago
10

A 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender​ selecti

on, the proportion of baby girls is greater than 0.5. Assume that sample data consists of 78 girls in 144 ​births, so the sample statistic of StartFraction 13 Over 24 EndFraction results in a z score that is 1 standard deviation above 0. Complete parts​ (a) through​ (h) below.
(a) Identify the null hypothesis and the alternative hypothesis. Choose the correct answer below.
a. H0:p=0.5 H1:P≠0.5.
b. H0:p≠0.5 H1:p=0.5.
c. H0:p=0.5 H1:p>0.5.
d. H0:p=0.5 H1:p<0.5.
(b) What is the value of α? (Type an integer or a decimal.)
(c) What is the sampling distribution of the sample statistic?
a. Normal distribution.
b. χ2.
c. Student (t) distribution.
(d) Is the test two-tailed, left-tailed, or right-tailed?
a. Right-tailed.
b. Left-tailed.
c. Two-tailed.
(e) What is the value of the test statistic? (Type an integer or a decimal).
(f) What is the P-value? (Round to four decimal places as needed.)
(g) What are the critical values(s)? (Round to three decimal places as needed. Use a comma to separate answers as needed.)
(h) What is the area of the critical region? (Round to two decimal places as needed.)
Mathematics
1 answer:
Andrews [41]3 years ago
7 0

Answer:

(a) Null Hypothesis, H_0 : p = 0.50

    Alternate Hypothesis, H_A : p > 0.50

(b) The value of level of significance (α) given in the question is 0.10.

(c) The sampling distribution of the sample statistic is Normal distribution.

(d) This test is right-tailed.

(e) The value of z test statistics is 0.96.

(f) The P-value is 0.1685.

(g) At 0.10 significance level the z table gives critical value of 1.282 for right-tailed test.

(h) We conclude that the proportion of baby girls is equal to 0.50.

Step-by-step explanation:

We are given that a 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender​ selection, the proportion of baby girls is greater than 0.5.

Assume that sample data consists of 78 girls in 144 ​births.

Let p = <u><em>population proportion of baby girls</em></u>

(a) So, Null Hypothesis, H_0 : p = 0.50     {means that the proportion of baby girls is equal to 0.50}

Alternate Hypothesis, H_A : p > 0.50     {means that the proportion of baby girls is greater than 0.50}

(b) The value of level of significance (α) given in the question is 0.10.

(c) The sampling distribution of the sample statistic is Normal distribution.

(d) This test is right-tailed as in the alternative hypothesis we are concerned for proportion of baby girls that is greater than 0.50.

(e) The test statistics that would be used here <u>One-sample z test for proportions</u>;

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

where, \hat p = sample proportion of baby girls =  \frac{78}{144} = 0.54

            n = sample of births = 144

So, <u><em>the test statistics</em></u>  =  \frac{0.54-0.50}{\sqrt{\frac{0.54(1-0.54)}{144} } }  

                                       =  0.96

The value of z test statistics is 0.96.

(f) <u>The P-value of the test statistics is given by;</u>

            P-value = P(Z > 0.96) = 1 - P(Z < 0.96)

                          = 1 - 0.8315 = 0.1685

<u></u>

(g) <u>Now, at 0.10 significance level the z table gives critical value of 1.282 for right-tailed test.</u>

(h) Since our test statistic is less than the critical value of z as 0.96 < 1.282, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region (which was to the right of value of 1.282) due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the proportion of baby girls is equal to 0.50.

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