it is stated that the standard deviation sample of the is 7.6 (n = 26). Test the claim to a significance level of 0.10 that the population standard deviation was greater than 7.0. Round all answers to 3 decimal places.
Null and alternative hypotheses:
Ho: The population standard deviation is = 7.0
Ha: The population standard deviation is > 7.0
Data given:
sample standard deviation = 7.6
level of significance = 0.10
n = 26
The test statistic used in making a decision is the chi-square.
Here's the formula:
chi-square = (sample size - 1)(sample variance)/(value specified in the null hypothesis)
Note: Convert both standard deviations to variances (7.0 and 7.6) when working with the formula. Remember that standard deviation is the square root of the variance; therefore, variance is standard deviation squared.
Degrees of freedom is equal to n - 1, which is 25.
Checking a chi-square table using significance level = 0.10 with 25 degrees of freedom, find the critical value. If the test statistic exceeds the critical value from the chi-square table, then the null is rejected. If the test statistic does not exceed the critical value from the table, then the null cannot be rejected and there is not enough evidence to support the claim that the population standard deviation is greater than 7.0 grams.
I hope this will help get you started.
The last line should read:
If the test statistic does not exceed the critical value from the table, then the null cannot be rejected and there is not enough evidence to support the claim that the population standard deviation is greater than 7.0. (The word "grams" should not have been included.)
Sorry for any confusion.
To test the claim, we'll use the chi-square distribution and the formula for the chi-square test statistic:
chi-square = (n - 1) * (sample standard deviation)^2 / (population standard deviation)^2
Given that the sample standard deviation (s) is 7.6, the population standard deviation (σ) is claimed to be greater than 7.0, and the sample size (n) is 26, we can calculate the chi-square test statistic as follows:
chi-square = (26 - 1) * (7.6)^2 / (7.0)^2
Calculating this expression:
chi-square = 25 * 57.76 / 49.00
chi-square = 28.90
Now, we need to compare this test statistic with the chi-square critical value at a significance level of 0.10 and degrees of freedom (df) equal to n - 1 = 26 - 1 = 25.
To find the critical value, we can refer to the chi-square distribution table or use statistical software. For a significance level of 0.10 with df = 25, the critical value is approximately 36.415.
Since the test statistic (chi-square = 28.90) is less than the critical value (36.415), we fail to reject the null hypothesis. Therefore, there is not enough evidence to support the claim that the population standard deviation is greater than 7.0 at a significance level of 0.10.