An online stock trading company makes part of their revenue from clients when th

Question

An online stock trading company makes part of their revenue from clients when the clients trade stocks therefore, it is important to the company to have an good idea of how many trades its clients are making in a given year. In a sample of 99 clients of an online stock trading company, the average number of trades per year was 87 with a standard deviation of 15. If you were to test the hypothesis that the average number of trades per year is greater than than the previous year when the average number of trades was 85 (using the 10% level of significance), what is the critical value? (please round your answer to 2 decimal places)

Explanation / Answer

Solution:-

In a sample of 99 clients , i.e., n = 99 of an online stock trading company, the average number of trades per year was 87 (x = 87) with a standard deviation of 15 (s = 15).

If you were to test the hypothesis that the average number of trades per year is greater than than the previous year when the average number of trades was 85 ( = 85) (using the 10% level of significance (0.10)), what is the critical value? (please round your answer to 2 decimal places)

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: >= 85
Alternative hypothesis: > 85 (claim)

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too large.

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) = 15 / sqrt(99) = 1.508
DF = n - 1 = 99 - 1 = 98
t = (x - ) / SE = (87 - 85) / 1.508 = 1.326

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.

Here is the logic of the analysis: Given the alternative hypothesis ( > 85), we want to know whether the observed sample mean is large enough to cause us to reject the null hypothesis.

The observed sample mean produced a t statistic test statistic of 1.326.

We use the t Distribution Calculator to find P(t < 1.326)

The P-Value is .093961.
The result is significant at p < .10

Interpret results. Since the P-value (0.09) is smaller than the significance level (0.10), we can reject the null hypothesis.

Conclusion. Reject the null hypothesis. We have sufficient evidence to prove the claim that the average number of trades per year is greater than the previous year when the average number of trades was 85.

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