Question 1
A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.
1. Formulate the hypotheses that can be used to determine whether the mean of all account balances is significantly different from $1,150.
Solution: Null hypothesis Hₒ: µ = $ 1150 against alternative H₁: µ≠$1150
2. Compute the test statistic.
Solution: since the sample size 81>30 therefore we use Z test and the test statistics will be
Z= sample standard deviation /√ sample ¿¿ sample mean − µ ¿
Under the null hypothesis the test statistics Z= 1200 – 1150 126 / √ 81 = 7.14
3. Using the p-value
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Sample 1 size 40 Sample 2 size 45 Sample 1 mean 72 Sample 2 mean 78 Sample 1 S.D. 6 Sample 2 S.D. 8 Degree of freedom = [s1^2/n1+s2^2/n2] ^2 / [(s1^2/n1) ^2/(n1-1) +(s2^2/n2) ^2/(n2-1)]
= [36/40 + 64/45] ^2 / [(36/40) ^2 /39+ (64/45) ^2/44] Degree of freedom 80.8018 Standard error of the difference between means = sqrt (s1^2/n1+s2^2/n2) = sqrt (6^2/40+8^2/45) = 1.523884 Confidence interval: (xbar1-xbar2) - t SE and (xbar1-xbar2) + t SE t (0.05,81) = 1.99 (72-78) - (1.99) (1.523884) and (72-78) + (1.99) (1.523884) 95% confidence interval is: (-9.03, -2.97)
2. At 95% confidence, use the p-value approach and test to determine if the average yearly income of marketing managers in the East is significantly different from the West. t = [ (x bar 1 – x bar 2) - d] / SE = -6/1.523884
Test statistic t = -3.937 (df = 80), p-value < .005, reject Hₒ and conclude that there is a significant
So, we should reject the null hypothesis H0. At a 0.05 level of significance level, we conclude that there is a significant difference between the average height for females and the average height for the males.
b) In order to calculate the mean or average for the governors and CEO’s, I added together all the figures and divided that sum by 4 since there
Select one (1) project from your working or educational environment that you would use the hypothesis test technique. Next, propose the hypothesis structure (e.g., the null hypothesis, data collection process, confidence interval, test statistics, reject or not reject the decision, etc.) for the business process of the selected project. Provide a rationale for your response.
2. For the following set of scores, fill in the cells. The mean is 74.13 and the standard deviation is 9.98.
To test the null hypothesis, if the P-Value of the test is less than 0.05 I will reject the null hypothesis.
In a simple random sample of students from colleges across the state of Washington, 490 are in favor decreasing the size of their college administration and 310 are not. Determine a 95% confidence interval estimate for the proportion of Washington college students who favor decreasing the size of their college administration.
So the p-value for this test is 0.006. Which is smaller than the significance level 0.05, as a result I am rejecting the null hypothesis based on the p-value approach as well.
The customers in this case study have complained that the bottling company provides less than the advertised sixteen ounces of product. They need to determine if there is enough evidence to conclude the soda bottles do not contain sixteen ounces. The sample size of sodas is 30 and has a mean of 14.9. The standard deviation is found to be 0.55. With these calculations and a confidence level of 95%, the confidence interval would be 0.2. There is a 95% certainty that the true population mean falls within the range of 14.7 to 15.1.
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
We conduct an independent sample t-test using Excel, and obtain the following output (see sheet T-TEST)
Since 79.38 is larger than 12.592, I can reject the null hypothesis and conclude this data to be statistically significant. My P-value is 0.00 and with a significance level of 0.05, I am able to reject the null hypothesis that income does not influence vote choice and I am able to conclude that income does have an effect on vote choice.
A pharmaceutical company is testing the effectiveness of a new drug for lowering cholesterol. As part of this trial, they wish to determine whether there is a difference between the effectiveness for women and for men. Using = .05, what is the value the test statistic?
The hypothesis of what is expected is that as a whole UMass Amherst college students will consume more alcohol than UMass Amherst recent graduates. In particular, the mean of the days per week alcohol is consumption and the mean of the amount of alcohol drank each day one drinks will be greater for UMass Amherst students than the means of UMass Amherst recent graduates. I am interested in calculating if the average amount of alcohol each time one drinks is exceptionally higher for those in college than not. Also, if the days per week correspond to the amount drank and if the mean of days per week and mean of amount drank will differ for students and/or alumni.
c.)Find a 95% confidence interval for the difference between the above obtained mean starting salaries.
We will calculate the averages of the samples and the general average and we will obtain: