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LenKa [72]
3 years ago
7

Perform the indicated goodness-of-fit test.

Mathematics
1 answer:
Alex17521 [72]3 years ago
8 0

Answer: Reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that the distribution of the sample is consistent with the distribution of the state populations.

Explanation:

The null and alternative hypotheses are:

H_{0}:The sample of 1000 subjects has a distribution that is consistent with the distribution of state populations.

H_{a}:The sample of 1000 subjects does not have a distribution that is consistent with the distribution of state populations.

Under the null hypothesis, the test statistic is:

\chi^{2}=\sum \frac{(O-E)^{2}}{E}

From the attachment, we clearly see the chi-square statistic is:

\chi^{2}=31.938

Now we have to find the chi-square critical value at 0.05 significance level for df = n - 1 = 4-1=3. Using the chi-square distribution table, we have:

\chi^{2}_{critical} =7.815

Since the chi-square statistic is greater than the chi-square critical value, we therefore reject the null hypothesis and there is sufficient evidence to warrant rejection of the claim that the distribution of the sample is consistent with the distribution of the state populations


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Answer:

Option b. should not be rejected

Step-by-step explanation:

We are given that the contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces.

We have to test whether the variance of the population is significantly more than 0.003, i.e.;

  Null Hypothesis, H_0 : \sigma = \sqrt{0.003}

Alternate Hypothesis, H_1 : \sigma > \sqrt{0.003}

The test statistics used here for testing variance is;

          T.S. = \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2}__n_-_1

where, s = sample standard deviation = 0.06

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So, Test statistics = \frac{(26-1)0.06^{2} }{0.003 } ~ \chi^{2}__2_5

                            = 30

So, at 5% level of significance chi square table gives critical value of 37.65 at 25 degree of freedom. Since our test statistics is less than the critical so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that null hypothesis should not be rejected and variance of population is 0.003.

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Step-by-step explanation:

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