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bezimeni [28]
3 years ago
9

Perform a hypothesis test using a 0.05 significance level to test Dr. Karabinus' claim (in the introductory video) about the suc

cess rate of the XSORT method of gender selection. Assume that the sample data that you are using consists of 55 girls born in 100 births.Part 1: What do we need to understand to construct a hypothesis test for Dr. Karabinus' claim? What values do we need to find to conduct a hypothesis test using a) the P-value method, b) the critical value method, and c) the confidence interval method?Part 2: Choose one of the three methods of hypothesis testing to test Dr. Karabinus' claim. Explain clearly how your hypothesis test is constructed and conducted without the use of technology. Show how the method works and give the meaning of the terminology in your own words.Part 3: Carefully word the final conclusion for your hypothesis test. Discuss the statistical significance of the results and your views regarding the use of gender selection methods?
Mathematics
1 answer:
viva [34]3 years ago
4 0

Answer:

a = 0.05

Claim: The success rate of the XSORT method of gender selection.

x = 55

n = 100

Null and alternative hypothesis is

H0:P=0.5 Vs P \neq0.5

p0 : 0.5

x : 55

n : 100

prop :\neq p0

calculate

we get

z = 1

p = 0.317310

so p value > 0.05 we fail to reject null hypothesis.

Conclusion : There is sufficient evidence that we support null hypothesis.

That is the success rate of the XSORT method of gender selection

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An insurance company selected a random sample of 500 clients under 18 years of age and found that 180 of them had had an acciden
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Answer:

a) The pooled proportion is p=0.3.

b) P-value = 0.000078

c) Lower bound = 0.0556

d) Upper bound = 0.1644

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the accident proportions differ between the two age groups .

Then, the null and alternative hypothesis are:

H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0

The significance level is 0.05.

The sample 1, of size n1=500 has a proportion of p1=0.36.

p_1=X_1/n_1=180/500=0.36

The sample 2, of size n2=600 has a proportion of p2=0.25.

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The difference between proportions is (p1-p2)=0.11.

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The pooled proportion, needed to calculate the standard error, is:

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The standard error for the difference between proportions can now be calculated as:

The estimated standard error of the difference between means is computed using the formula:

s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.3*0.7}{500}+\dfrac{0.3*0.7}{600}}\\\\\\s_{p1-p2}=\sqrt{0.00042+0.00035}=\sqrt{0.00077}=0.0277

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z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.11-0}{0.0277}=\dfrac{0.11}{0.0277}=3.964

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

P-value=2\cdot P(t>3.964)=0.000078

As the P-value (0.000078) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that the accident proportions differ between the two age groups.

If we want to calculate the bounds of a 95% confidence interval, we start by calculating the margin of error.

For a 95% CI, the critical value for z is z=1.96.

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MOE=z \cdot s_{p1-p2}=1.96\cdot 0.0277=0.0544

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LL=(p_1-p_2)-z\cdot s_{p1-p2} = 0.11-0.0544=0.05561\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= 0.11+0.0544=0.16439

The  95% confidence interval for the population mean is (0.0556, 0.1644).

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Answer:

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