Part B Course Project Math 533 Essay

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).

2) Compute the standard deviation for each of the four samples. Does the assumption of .21 for the population standard deviation appear reasonable?

We know that +/- 1.96 standard deviations from the mean will contain 95% of the values. So, we can get the standard deviation by:

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:

Based on the results of the hypothesis tests, both p-value approach and critical value approach, corrective action should be taken for sample 3. Samples 1,2 and 4 provide evidence that we cannot reject Ho, and therefore the

A business wants to estimate the true mean annual income of its customers.  It randomly samples 220 of its customers.  The mean annual income was $61,400 with a standard deviation of $2,200.  Find a 95% confidence interval for the true mean annual income of the business’ customers.

Because it is one sided, α will remain .05. Thus, comparing .0384 to .05, it is lower. In conclusion, the mean age of stars is indeed greater than 3.3. We reject the null hypothesis.

P-value represents a decimal between 1.0 to below .01. Unfortunately, the level of commonly accepted p-value is 0.05. The level of frequency of P>0.05 means that there is one in twenty chance that the whole study is just accidental. In other words, that there is one in twenty chance that a result may be positive in spite of having no actual relationship. This value is an estimate of the probability that the result has occurred by statistical accident. Thus, a small value of P represents a high level of statistical significance and vice

To test the null hypothesis, if the P-Value of the test is less than 0.05 I will reject the null hypothesis.

Fry Brothers heating and Air Conditioning, Inc. employs Larry Clark and George Murnen to make service calls to repair furnaces and air conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. Assume the population standard deviation for Larry Clark is 1.05 calls per day and 1.23 calls per day for George Murnen. A random sample of 40 days last year showed that Larry Clark made an average of

With a P-value of 0.00, we have a strong level of significance. No additional information is needed to ensure that the data given is accurate.

Since the P value from the two tail test is 0.0002720 this means the data is statistically significant and we are not incorrectly rejecting a true null hypothesis.

After using the sample of 200 students in the BS program at Whatsamatta U. I conducted a hypothesis test using Analysis of Variance to determine if there is a difference in the mean GPA, for those who are unemployed vs. working part time vs. working full time. When conducting this test I will use a significance level of .05. After loading the data the summary table reveals that the average GPA of students who are unemployed is 3.388. The average of the students who work part time is 3.49. And the average GPA of the students who work full time is 3.563. After loading my data, I discovered that not all of my means are equal in the 3 quiz format. With a p-value of 0.130, I will reject the null hypothesis. Since my probability value is within my

Test Statistic, t: -2.6515, Critical t: ±2.146376, P-Value: 0.0191, Degrees of freedom: 14, 95% Confidence interval: -2.895174 < µ1-µ2 < -0.3048258

Using a 10% (0.1) significant level also known as Alpha, the p-value of the two tailed test shows that it is 38% (0.38) which is obviously higher than 10% which implies that we cannot reject the null hypothesis. Ho: μ1 = μ2