The plant manager believes that the operators take a lunch break for more than 60 minutes. A sample of 15 operators from a normal population were randomly selected, and it was found out that their mean lunch break is 60 minutes with a standard deviation of 5 minutes. At 0.05 level of significance, is the mean lunch break greater than 60 minutes?
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A: The objective of this question is to test the claim that the mean number of customers who make a…
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Q: A new small business wants to know if its current radio advertising is effective. The owners decide…
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- A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 10 days following no advertisements, the mean was 18.3 purchasing customers with a standard deviation of 1.8 customers. On 7 days following advertising, the mean was 19.4 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.02 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 3 of 3: Draw a…A pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 44 participants in the treatment group lowered their cholesterol levels by a mean of 18.7 points with a standard deviation of 3.3 points. The 39 participants in the control group lowered their cholesterol levels by a mean of 18.1 points with a standard deviation of 2.1 points. Assume that the population variances are not equal and test the company’s claim at the 0.02 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha : μ1⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯μ2 Step 2 of 3: what is the test statistic Step 3 of 3: draw a conclusion, fail or reject. Is…A public bus company official claims that the mean waiting time for Bus # 14 during peak hours is approximately 10 minutes. Karen took Bus # 14 during peak hours on 36 different occasions. Her mean waiting time was 8.7 minutes. Assume that the population standard deviation o of 2.9 minutes is known. At the 0.01 significance level, test if the mean of all the peak hours waiting time for Bus # 14 is significantly different from 10 minutes.
- A local community college reported that the average SAT score of its students is 960 with a standard deviation of 82. If a student scored 975 on the SAT, what is the student's z score? Did the student do better or worse on the SAT compared to the average student in the college?Suppose that a report by a leading medical organization claims that the healthy human heart beats an average of 72 times per minute. Advances in science have led some researchers to question if the healthy human heart beats an entirely different amount of time, on average, per minute. They obtain pulse rate data from a sample of 85 healthy adults and find the average number of heart beats per minute to be 76, with a standard deviation of 13. Before conducting a statistical test of significance, this outcome needs to be converted to a standard score, or a test statistic. What would that test statistic be? (Use decimal notation. Give your answer to one decimal place.) test statistic:In a study of facial behavior, people in a control group are timed for eye contact for a 5 minute period. They are normally distributed with a mean of 177 seconds and a standard deviation of 53 seconds. I need to find the percentage of times within 53 seconds of the mean of 177 seconds and round it one decimal place if needed.
- According to the Insurance Association in US, the average annual expenditure for automobile insurance is $605. Suppose that in a simple random sample of 80 residents of New York, for most recent year their average expenditure was $638.62 with the standard deviation of $106.05. Based on these data, examine whether the average annual insurance for motorists in New York might be different from that in national level. Using the 0.05 level of significance, what conclusion do you reach? Also, construct and interpret 95% confidence interval for the population mean. Is the hypothesized mean within the interval? Is this consistent with the findings of the hypothesis test conducted?A pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 34 participants in the treatment group lowered their cholesterol levels by a mean of 22.2 points with a standard deviation of 3.4 points. The 42 participants in the control group lowered their cholesterol levels by a mean of 21.2 points with a standard deviation of 1.8 points. Assume that the population variances are not equal and test the company's claim at the 0.10 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho:₁ = ₂ Ha: M •M₂In 16 women with diabetes, the morning blood sugar mean was 49.4 with a standard deviation of 24.0. In normal healthy women, the mean was 47.0. From the sample information, can we conclude that the blood sugar mean in women with rheumatoid arthritis is greater than that in normal women? Assume a significance level of 0.05.
- A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 13 days following no advertisements, the mean was 23.9 purchasing customers with a standard deviation of 1.9 customers. On 6 days following advertising, the mean was 24.7 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.01 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 3 of 3: Draw a…Steve believes that his wife's cell phone battery does not last as long as his cell phone battery. On eight different occasions, he measured the length of time his cell phone battery lasted and calculated that the mean was 19.7 hours with a standard deviation of 9.4 hours. He measured the length of time his wife's cell phone battery lasted on twelve different occasions and calculated a mean of 18.9 hours with a standard deviation of 5.3 hours. Assume that the population variances are the same. Let Population 1 be the battery life of Steve's cell phone and Population 2 be the battery life of his wife's cell phone. Step 1 of 2: Construct a 90 % confidence interval for the true difference in mean battery life between Steve's cell phone and his wife's. Round the endpoints of the interval to one decimal place, if necessary.In a random sample of 6 cell phones, the mean full retail price was 517.30 and standard deviation is 184.00. Further research suggests that the population mean is 425.73. Does the t-value for the original sample fall between -t0.99 and t0.99?