Respuesta :
In this question, the variance of the population is tested. From the data given in the exercise, we build the hypothesis, then we find the value of test statistic and it's respective p-value, to conclude the test. From this, it is found that the conclusion is:
The p-value of the test is 0.0038 < 0.05, which means that the investor can conclude that the variance of the share price of the bond fund is different than claimed at α = 0.05.
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Claimed variance of 0.19:
This means that at the null hypothesis, it is tested if the variance is of 0.19, that is:
[tex]H_0: \sigma^2 = 0.19[/tex]
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Test if the variance of the share price of the bond fund is different than claimed at α = 0.05.
At the alternative hypothesis, it is tested if the variance is different of the claimed value of 0.19, that is:
[tex]H_1: \sigma^2 \neq 0.19[/tex]
The test statistic for the population standard deviation/variance is:[tex]\chi^2 = \frac{n-1}{\sigma_0^2}s^2[/tex]
In which n is the sample size, is the value tested for the variance and s is the sample standard deviation.
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0.19 is tested at the null hypothesis, as the variance:
This means that [tex]\sigma_0^2 = 0.19[/tex]
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He takes a random sample of the share prices for 22 days throughout the last year and finds that the standard deviation of the share price is 0.2517.
This means that [tex]n = 22, s^2 = (0.2517)^2 = 0.0634[/tex]
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Value of the test statistic:
[tex]\chi^2 = \frac{n-1}{\sigma_0^2}s^2[/tex]
[tex]\chi^2 = \frac{21*0.0634}{0.19}[/tex]
[tex]\chi^2 = 7[/tex]
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P-value of the test and decision:
The p-value of the test is found using a chi-square for the variance calculator, considering a test statistic of [tex]\chi^2 = 7[/tex] and 22 - 1 = 21 degrees of freedom, and a two-tailed test(test if the mean is different of a value).
Using the calculator, the p-value of the test is 0.0038.
The p-value of the test is 0.0038 < 0.05, which means that the investor can conclude that the variance of the share price of the bond fund is different than claimed at α = 0.05.
For more on hypothesis tests using variances/standard deviation, you can check https://brainly.com/question/13993951
