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Testing Random Samples

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A) An iron foundry asserts that the mean weight of the castings it produces is 20.0 kg. The weights of a sample of 10- castings are:

19.8 20.3 20.6 21.1 19.3
19.6 20.1 20.8 21.1 21.3

Assuming that the weight of a casting has an approximately normal distribution, test whether or not the assertion is correct.

b) The police are increasingly concerned about the fact that large numbers of 999 calls are not true emergencies. They are contemplating using an advertising campaign to warn of the consequences of abusing the 999 number. Because of the cost of such a campaign, it can only be justified if more that 25% of all 999 calls are not emergencies. A random sample of 200 recent 999 calls is selected and it is determined that 66 were non-emergencies. Does this data support going ahead with the advertising campaign?

c) A students union uses the profits from a snack dispenser to help finance its activities. The price per snack had been 40 pounds for a long time, and the average daily revenue during that period had been 50.00 pounds. The price was recently increased to 45 pounds per snack. A random sample of the 20 days subsequent to the price increase showed a sample average revenue of 47.30 pounds with a standard deviation of 4.20 pounds. Does this data suggest that the true average daily revenue has decreased from its value prior to the price increase?

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Solution Summary

This solution provides a null and alternative hypothesis for all cases and conducts a statistical test to determine the p-value. It also decides weather or not to accept or reject the null hypothesis. All steps are shown with explanations.

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a) An iron foundry asserts that the mean weight of the castings it produces is 20.0kg. The weights of a sample of 10- castings are:

19.8 20.3 20.6 21.1 19.3
19.6 20.1 20.8 21.1 21.3

Assuming that the weight of a casting has an approximately normal distribution, test whether or not the assertion is correct.

Solution. We need to set up hypotheses

H0:
H1:

Using Excel, we can get the sample mean and sample standard deviation as follows:

Column1

Mean 20.4
Standard Error 0.218581
Median 20.45
Mode 21.1
Standard Deviation 0.691215
Sample Variance 0.477778
Kurtosis -1.30362
Skewness -0.25234
Range 2
Minimum 19.3
Maximum 21.3
Sum 204
Count 10

So,

To compute the test t-statistic

Note that the degree ...

Solution provided by:
Changping Wang, MA
Education
  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
  • "Thank you"
  • "Thank you very much for your valuable time and assistance!"
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