please explain how you got answers I am having a really hard time with this chap

Question

please explain how you got answers I am having a really hard time with this chapter. Thanks!

1.what is the value of the standarized statistic?(round to two decimal points)

2. what is(are) the critical values? (round to three decimal points)

3. decide whether to reject of fail to reject the null hypothesis.

Use a t-test to test the claim about the population mean H at the given level of significance a using the given sample statistics. Assume the population is normally distributed. Claim: -51,900; a 0.10 Sample statistics: x 53,108, s -1800, n -19 AE Click the icon to view the t-distribution table. What are the null and alternative hypotheses? Choose the correct answer below. O A. Ho: 51,900 O B. Ho: 51,900 Ha: 51,900 Ha 51,900 O C. Ho us 51,900 O D. Ho: 51,900 Ha: 51,900 Ha: u 51,900

Explanation / Answer

Solution:-

Given,
Population is normally distributed,
Claim : = 51,900
alpha = 0.10
Sample statistics : xbar = 53,108
s = 1800
n = 19

For the null and alternate hypothesis option (D) is correct, because we never choose equals to sign for alternate hypothesis, either it is, not equal to sign or </> sign for alternate hypothesis.

Now lets work for the t - test, we will use hypothesis test for mean,

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: = 51,900
Alternative hypothesis: 51,900

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n) = 1800 / sqrt(19) = 412.95

DF = n - 1 = 19 - 1 = 18
t = (x - ) / SE = (53,108 - 51,900)/412.95 = 2.93

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size..

We use the t Distribution Calculator to find P(t < 2.93)

The P-Value is 0.008946.
The result is significant at p < 0.10.

Interpret results. Since the P-value (0.008946) is smaller than the significance level (0.10), we can reject the null hypothesis.

Conclusion. Reject the null hypothesis. We have insufficient evidence to Claim that = 51,900.

From the above :-

1. Value of the standarized statistic = 2.93

2. Critical values using z table for alpha 0.10 is -1.2816 to 1.2816.

3. Reject the null hypothesis.

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