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In 1990, the mean height of women 20 years of age or older was 63.7 inches based on data obtained from the CDC. Suppose that a random sample of 45 women who are 20 years of age or older in 2015 results in a mean height of 63.9 inches with a standard deviation of 0.5 inch. Does your sample provide sufficient evidence that women today are taller than in 1990? Perform the appropriate test at the 0.05 level of significance

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The sample provides sufficient evidence to suggest that women today are taller than in 1990. The one-sample t-test shows that the mean height of women in the 2015 sample (63.9 inches) is statistically significantly greater than the mean height of women in 1990 (63.7 inches) at the 0.05 level of significance.

To determine if the sample provides sufficient evidence that women today are taller than in 1990, we need to perform a hypothesis test. The null hypothesis (H0) assumes that there is no difference in the mean height of women between 1990 and 2015. The alternative hypothesis (Ha) assumes that the mean height of women in 2015 is taller than in 1990.

To perform the appropriate test, we can use a one-sample t-test. Here are the steps:

1. Set up the hypotheses:

- H0: The mean height of women in 2015 is the same as in 1990 (μ = 63.7).

- Ha: The mean height of women in 2015 is taller than in 1990 (μ > 63.7).

2. Determine the significance level (α):

- The significance level given in the question is 0.05. This means we want to be 95% confident in our conclusion.

3. Calculate the test statistic:

- The test statistic for a one-sample t-test is calculated as: t = (sample mean - population mean) / (sample standard deviation / √n).

- In this case, the sample mean is 63.9, the population mean is 63.7, the sample standard deviation is 0.5, and the sample size is 45.

- Plugging in the values, we get: t = (63.9 - 63.7) / (0.5 / √45).

4. Determine the critical value:

- Since the alternative hypothesis is one-sided (μ > 63.7), we need to find the critical value from the t-distribution with (n-1) degrees of freedom and the chosen significance level.

- In this case, the degrees of freedom is 44 (n-1), and the chosen significance level is 0.05.

- Using a t-table or a t-distribution calculator, we find the critical value to be approximately 1.677.

5. Make a decision:

- If the test statistic is greater than the critical value, we reject the null hypothesis. This means there is sufficient evidence to conclude that women today are taller than in 1990.

- If the test statistic is less than or equal to the critical value, we fail to reject the null hypothesis. This means there is not sufficient evidence to conclude that women today are taller than in 1990.

6. Calculate the test statistic and compare it to the critical value:

- Plugging in the values, we get: t = (63.9 - 63.7) / (0.5 / √45) ≈ 1.897.

- Since the test statistic (1.897) is greater than the critical value (1.677), we reject the null hypothesis.

7. Make a conclusion:

- Based on the results of the hypothesis test, there is sufficient evidence to conclude that women today are taller than in 1990.

In summary, the sample provides sufficient evidence to suggest that women today are taller than in 1990. The one-sample t-test shows that the mean height of women in the 2015 sample (63.9 inches) is statistically significantly greater than the mean height of women in 1990 (63.7 inches) at the 0.05 level of significance.

User Jdnichollsc
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2 votes

Answer:

We reject H₀ , we have enough evidence, women today are taller than in 1990

Explanation:

The problem is one tail test. We assume height of women follows normal distribution with σ = 0,5 in.

Then null hypothesis : H₀ ⇒ μ₀ = 63.7

Alternative hipothesis: Hₐ ⇒ μ > 63.7

Significance level 0,05 ⇒ z(c) = 1.645

Statistic z :

z (c) = ( μ - μ₀ ) / (0,5/√45) ⇒ z(c) = (63.9 - 63.7 ) / (0,5 / √45)

z(c) =( 0,2* √45 ) / 0,5

z(c) = 2,68

Evaluation : z(s) > z(c) 2.68 > 1.645

z(c) is in the rejection area we reject H₀

User Ajitha
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