Question

#1) Scientists are working on a synthetic vaccine​ (antigen) for a particular parasite. The success of the vaccine hinges on the characteristics of DNA in peptide​ (protein) produced by the antigen. In one​ study, scientists tested alleles of​ antigen-produced protein for level of peptide. For a sample of 4 ​alleles, the mean peptide score was 1.43 and the standard deviation was 0.11.

a. Use this information to construct a 90​%

confidence interval for the true mean peptide score in alleles of the​ antigen-produced protein.

b. Interpret the interval for the scientists.

c. What is meant by the phrase

​"90​% confidence?"

#2 ) In a random sample of

sixsix

microwave​ ovens, the mean repair cost was

​$70.0070.00

and the standard deviation was

​$14.0014.00.

Assume the population is normally distributed and use a​ t-distribution to construct a

95​% confidence interval for the population mean μ.

What is the margin of error of μ​?Interpret the results.

a. What is the margin of error of μ​? Interpret the results.

b. The margin of error is ____

#3)

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Which interval is​ wider? If​ convenient, use technology to construct the confidence intervals.

A random sample of 32 gas grills has a mean price of $628.50.

Assume the population standard deviation is $56.50

a. The​ 90% confidence interval is ?

b. The​ 95% confidence interval is?

Answer 1

Solution-1: Sample size-n 4 Sample mean score 1.43 Sample Standard Deviation =s-0.11 For 90% confidence interval, 1-0.90-0.10 a/2 0.10/2 0.05 Degrees of freedom-df n 1 4 1 3 From t-distribution table, the value of t having an area of (a/2 0.05) in its upper tail and for (df to.5 2.353 3) is given by Margin of Error: 0.11 MoE 0.129 Part-A: 90% confidence interval: Lower Limit of 90% confidence interval for mean peptide score μ is = 1.43-0.129 = 1.301 Upper Limit of 90% confidence interval for mean peptide score is-1.43 + 0.129 1.559

Part-B: Interpretation: we are 90% confident that population mean peptide score lies between 1.301 and 1.559 Part-C: 90% confidence means that if we take repeated samples from this population and measure the mean peptide score for each of the sample, then 90% of the samples are expected to have mean peptide score between 1.301 and 1.559 Solution-2: Sample size-n- 6 Sample mean repair cost =x = $70 Sample Standard Deviation =s = $14 For 95% confidence interval, α 1-0.95-0.05 α/2 = 0.05/2= 0.025 Degrees of freedom =df = n-1 = 6-1 5 From t-distribution table, the value of t having an area of (a/2 to.025 2.571 0.025) in its upper tail and for (df 5) is given by Part-A: Margin of Error: MoE t0.025 x s/Vn 14 V6 MoE = 14.692

Part-B: 95% confidence interval: Lower Limit of 95% confidence interval for population mean fl ls-70-14.692-55.308 Upper Limit of 95% confidence interval for population mean μ is-70 + 14.692-84.692 we are 95% confident that population mean repair cost for microwave ovens lies between $55.308 and $84.692 Solution-3: Sample sizen 32 Sample mean price $628.50 Population Standard Deviation-σ-$56.50 Part-A: For 90% confidence interval, α-1-0.90-0.10 a/2 0.10/2-0.05 From normal distribution table, the value of z having an area of (a/2 0.05) in its upper tail is given by Zo.05 1.645 56.50 MoE1.645 x 32 MoE 16.429

90% confidence interval: Lower Limit of 90% confidence interval for population mean μ is-628.50-16.429-612.071 Upper Limit of 90% confidence interval for population mean μ is-628.50 + 16.429-644.929 Part-B: For 95% confidence intervalα-1-0.95-0.05 a/2 0.05/2 0.025 From normal distribution table, the value of z having an area of (a/2 0.025) in its upper tail is given by Z0.025 1.96 , MoE = z0.025 56.50 V32 → MOE = 19.576 95% confidence interval: Lower Limit of 95% confidence interval for population mean μ is = 628.50-19.576 = 608.924 Upper Limit of 95% confidence interval for population mean μ is-628.50 + 19.576 648.076