Essay about Final Exam Math 302 Amu

From the above output, we can see that the p-value is 0.000186, which is smaller than 0.05 (if we select a 0.05 significance level).

3. Write the answer to the following problems in the correct number of significant figures.

2. In order to determine the average amount spent in November on Amazon.com a random sample of 144 Amazon accounts were selected. The sample mean amount spent in November was $250 with a standard deviation of $25. Assuming that the population standard deviation is unknown, what is a 95% confidence interval for the population mean amount spent on Amazon.com in November?

We reject Ho if χ2 > χα2. At α=0.05, with 4 degrees of freedom, the critical value becomes χα2=9.488 (table E.4)

15 In testing the hypotheses: H0 β1 ’ 0: vs. H1: β 1 ≠ 0 , the following statistics are available: n = 10, b0 = 1.8, b1 = 2.45, and Sb1= 1.20. The value of the test statistic is:

(b) The value of Z with an area of 5% in the right tail, but not the sample mean.

The mean for the column “mean” is 3.56. It is very close to the parameter of interest but is not equal to it. You can calculate a confidence interval for the mean of the mean column, but a specific confidence interval would need to be provided. In that case, the confidence interval would be centered on 3.56, not 3.5.

Because the p-value of .035 is less than the significance level of .05, I will reject the null hypothesis at 5% level.

7. The data set for this problem can be found through the Pearson Materials in the Student Textbook Resource Access link,

The next table shows the results of this independent t-test. At the .05 significance level, can we conclude

Develop 95% confidence intervals for the proportion of subscribers who have broadband access at home and proportion of subscribers who have children.

The second step is defining the significance level, determining the degrees of freedom and finding the critical value. The a-level shows that for a result to be statistically significant, it cannot occur more than the a-level percentage of time by chance. The critical value can be obtained by using the t-test table. The degrees of freedom is

2.52 0.63 49.19 3.50 74.31 39.83 69.86 96.97 5.34 1.46 0.65 0.57 5.73 1.47 35.22% 13.06% 7.01% 12.42% 29.41%

Since 3.27 the t statistic is in the rejection area to the right of =1.701, the level of

5) From calculations, computed z value is more than -1.65 and falls within Ho not rejected region. Ho is not rejected at α = 0.05 & α = 0.01 significance levels.