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A factory produces a component used in manufacturing computers. A sample of components is tested prior to shipment to determine how many defective components it contains. In a random sample of 367 components 65 were found to be defective. Using the method described in the Section 6.1 lecture notes, find the lower limit for the 95% confidence interval for the true proportion of defective components produced by the factory. Give your answer correct to 2 decimal places. Answer:
A factory produces a component used in manufacturing computers. The defect rate is supposed to be at most 0.05 (5%). A sample of components is tested prior to shipment to determine how many defective components it contains. In a random sample of 500 components, 35 were found to be defective. We are testing the following pair of hypotheses for p, the true defect rate: Ho: p = 0.05 Ha: p > 0.05 We are using (alpha) a = 0.01. What is the P-value of this test? You may want to use something other than a hand-held calculator for this as there are multiple steps and we want to avoid rounding error. For an example of a similar calculation in Excel see here. Select one: O a. 0.020 O b. 0.980 O c. 0.010 O d. 0.004
Suppose that we are interested in how much money the average cricket supporter spends on food at a single cricket match. To try to answer this, we asked 42 randomly selected supporters at a cricket match how much money they had spent on food. The sampled results show that the sample mean and standard deviation were NZ$45.00 and NZ$6.36, respectively. Use the sample statistics to calculate an estimate of the standard error for the sample mean. Hint: We show how to estimate a standard error for a sample mean in Section 6.2. Give your answer to 2 nal places. Answer:
Suppose a confidence interval for a population mean turns out to be (1,000, 2,100) at the 80% confidence level. Suppose we repeat our calculation with a different sample size and confidence level. Which of the following changes we would expect to result in an increased interval width (select the option for 'None of these' if you do not think any of the suggested changes can be expected to give an increased confidence interval width). Select one: a. Changing the confidence level to 60%. b. Changing the confidence level to 90%. C. None of these. O d. Increasing the sample size.
You manage to obtain data on 17 recently resold 5 year old foreign sedans of that model. These 17 cars were resold at an average price of $12,240 with a standard deviation of $650. Assuming that these car resale values follow an approximately normal distribution, calculate a 95% confidence interval for the true mean resale value of 5 year old foreign sedans. S We are standardising # with so we will need to use the t distribution to calculate t* Vn (rather than using the Standard Normal distribution to calculate z*). You can calculate t* using the Theoretical Distributions - t - function in StatKey. You will need to know the 'degrees of freedom'. Degrees of freedom = n - 1. Select one: 12,240 + 333 12,240 + 309 12,240 + 334 12,240 + 345
A discount airline allows staff 34.0 minutes between flights to clean the aircraft interior. The aircraft workers union claims that that average time that it takes to this cleaning is more than 34.0 minutes. They want to carry out a hypothesis test. They take a random sample n = 39 flights and time how long it takes to carry out the cleaning. They find that the sample mean is 37.52 minutes and the sample standard deviation is 4.37 minutes. Calculate the test statistic for this test and enter your answer below. Give your answer to 2 decimal places. This will be a t-test statistic because we are standardising the sample mean using an estimate for the standard error (based on s, the sample standard deviation). You can see an example of a similar calculation done in Excel here. Answer:
A jelly-baby manufacturer is aware that many people like the red jelly babies best, but wants to know if this is the same in the North and South Island. From a survey, we have the following data: North Islanders: Of 350 North Islanders surveyed, 240 prefer red jelly babies. South Islanders: Of 180 South Islanders surveyed, 70 prefer red jelly babies. Say that the true proportion of North Islanders who prefer red jelly babies is p1 and the true proportion of South Islanders who prefer red jelly babies is P2. If our null hypothesis is Ho: P1 - P2 = 0 and our alternative hypothesis is Ha: p1 - P2 = 0, what is the z test statistic for testing the difference P1 - P2? = Calculate this z test statistic and enter your answer, correct to 3 decimal places, into the box below. You can download an example of a similar calculation being done in Excel here. Note that there are quite a few steps to calculating a z test statistic for two proportions and it's important to avoid rounding error at each step. Answer: