Question 1An independent researcher wants to test a drug companies claims that their leading brand of pain relief works in less than 10 minutes. A random sample of 9 people was taken under clinical conditions and the results of time for pain relief to occur (in minutes) can be summarized as follows: 11.810.29.48.710.89.712.59.510.7Calculate a 95% confidence interval for the average time taken for the drug to take effect. Justify your choice of distribution for Mean(x) = 11.8+10.2+9.4+8.7+10.8+9.7+12.5+9.5+10.7=93.3/9+ 10.4Standard deviation= (11.8-10.4)^2 +(10.2-10.4)^2 +(9.4-10.4)^2 +(8.7-10.4)^2 +(10.8-10.4)^2 +(9.7-12.5)^2 +(12.5-10.4)^2 +(9.5-10.4)^2 +(10.7-10.4)^2= 11.82 =3.442 Z score = 0.4/1.147= 0.348Confidence interval = 95%, critical value= 0.4750=1.96Interpret the confidence interval (Max.
100 words)The confidence interval indicates the level of uncertainty arising from the use of a particular sampling method. A 95 percent confidence interval implies that there is a 95 percent possibility for interval estimate to have the population parameter included. The Z score is less than the critical value and this implies that the null hypothesis should be rejected. The drug does not relieve pain in less than the indicated minutes. The researcher is thinking of alerting authorities over misleading advertising for the pain reliever.
From the confidence interval calculated above, is there sufficient evidence for the researcher to take action? Since the null hypothesis should be rejected, the researcher can take substantial action against the medical firm. This is because, the research depicts that the drug does not work within the provided time frame of 10 minutes. Question 2In a survey of 1100 children, it was found that 715 played computer games for more than 1 hour per day. Calculate a 90% confidence interval for the proportion of children who played computer games for more than 1 hour per day. 715/ 1100= 21.56Confidence interval = 90%, critical value= 1.65Interpret the confidence intervalThe null hypothesis should be accepted as it is within the 90 percent confidence intervalQuestion 3A warehouse style retailer is conducting a marketing survey in the emerging area of Flowerdale for potential new store openings.
Due to the size of investment required, they want to be certain that there is sufficient demand and income for their products in the area. What sample size is needed if the company wants to be 99% confident of estimating the true population average income to within $3,000?
Census data indicates that income standard deviation for Flowerdale is $12,000. S/ Nb) What sample size is needed to estimate the population proportion of customers willing to buy the products to within ±0.025 with 95% confidence? Question 4A company wishing to produce ammunition for clay-target shooting must ensure consistent weights of rounds to comply with Olympic standards. Standards indicate that the round should way 28 grams. A recent sample of 40 shotgun rounds manufactured had an average weight of 28.2 grams.
The recent audit of machine accuracy determined the population standard deviation of shell weights was 0.5 grams. Is there sufficient evidence that the mean weight of shotgun shells is different to the Olympic standards? Test the claims where α = 0.05 and state any assumptions used in the analysis. Z score= 0.2/0.06=2.59Confidence interval = 0.95, the critical value= 1.96The Z score is more than the critical value and hence the null hypothesis should be accepted. There is sufficient evidence that the mean weight of shotgun shells is different to the Olympic standards.