McGraw hill unit 3 practice problem answers sixth grade lesson 3-1 MATH

Mathematics

2 answers:

Anna007 [38]3 years ago

6 0

Answer:

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Step-by-step explanation:

Anna71 [15]3 years ago

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Post the questions please

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Perform the indicated goodness-of-fit test.

Alex17521 [72]

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