ASK YOUR TEACHERWhat price do farmers get for their watermelon crops? In the third week of July, a random sample of 41 farming regions gave a sample mean of x = $6.88 per 100 pounds ofwatermelon. Assume that o is known to be $1.94 per 100 pounds.(a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop (in dollars). What is the margin oferror (in dollars)? (For each answer, enter a number. Round your answers to two decimal places.)lower limit$4upper limitmargin of error(b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.35 for the mean price per 100 pounds of watermelon. (Enter a number. Roundup to the nearest whole number.)farming regions(c) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop (in dollars). What is the margin of error (indollars)? Hint: 1 ton is 2000 pounds. (For each answer, enter a number. Round your answers to two decimal places.)lower limit2$upper limitmargin of error%24

Given : a) Given : Sample Size , n= 41 Sample Mean , x̄ =6.88 population standard deviation , σ =1.94 confidence level,c=0.9solution : Significance level, α =1-0.9 = 0.1 critical value at α=0.1 is Zα/2 = 1.645 ....from the standard normal distribution table Margin of error , ME = Zα/2 * σ / √ n = 1.645x1.94/ √41 = 0.498397 = 0.5 90% confidence interval =6.88±1.645x1.94/ √41 =6.88 ±0.498397 =(6.88 - 0.498397, 6.88 +0.498397) =(6.38 , 7.38) lower limit = 6.38 upper limit = 7.38 Margin of error = 0.5 b) Given : Margin of error, ME = 0.35 population standard deviation, σ = 1.94 confidence level,c= 0.9 Significance level, α =1- 0.9 = 0.1 critical value, critical value at α= 0.1 is Zα/2 = 1.645 .....from standard normal distribution table formula for the minimum sample size needed to estimate a population mean to within ME units is =( 1.645 )^2 x ( 1.94 )^2/( 0.35 )^2 = 83.1379 ( Note : since it is impossible to take a fractional observation we round sample size to next whole number ) n = 84 minimum sample size required is 84c) Given : Sample Size , n= 15 Sample Mean , x̄ =6.88 population standard deviation , σ =1.94 confidence level,c=0.9 Significance level, α =1-0.9 = 0.1 critical value at α=0.1 is Zα/2 = 1.645 ....from the standard normal distribution table Margin of error , ME = Zα/2 * σ / √ n = 1.645x1.94/ √15 = 0.82 90% confidence interval =6.88±1.645x1.94/ √15 =6.88 ±0.82399 =(6.88 - 0.82399, 6.88 +0.82399) =(6.06 , 7.7) lower limit = 6.06 upper limit = 7.7 Margin of error = 0.82