A market researcher for a consumer electronics company wants to study the media viewing behavior of residence of a particular area. A random sample of 40 respondents is selected and each respondent is instructed to keep a detailed Record of time spent in gaged doing contact across all screens in a particular week. The results are as follows:continuing time per week: x=54 hours, s=4.1 hours. 8 respondents have high definition on at least one television set. hour assuming that the population standard deviation is equal to 5 hours?(Type a whole number.)d. How many respondents need to be selected to be 90% confident of being within ±0.09 of the population proportion who have HD on at least one tele(Type a whole number.)e. Based on (c) and (d), how many respondents should the market researcher select if a single survey is being conducted?(Type a whole number.)MoreF3F2escF1@#$234 .....a. Construct a 90% confidence interval estimate for the mean content viewing time per week in this area.SHS(Round to two decimal places as needed.)b. Construct a 90% confidence interval estimate for the population proportion of residents who have HD on at least one television set.(Round to four decimal places as needed.)c. Suppose that the market researcher wants to take another survey in a different location. What sample size is required to be 90% confident of estimating the population mean content viewing time to within ± 1hour assuming that the population standard deviation is equal to 5 hours?(Type a whole number.)d. How many respondents need to be selected to be 90% confident of being within ±0.09 of the population proportion who have HD on at least one television set if no previous estimate is available?(Type a whole number.)(1,1)MoreF2escF1@#2341QWERGH.

Note: "Since you have posted a question with multiple sub -parts, we will solve first three sub parts for you. To get remaining subpart solved please re post the complete question and mention the sub parts to be solved". t-distribution should be used if the population standard deviation is unknown z- distribution should be used if the population standard deviation is known. Confidence interval is a range of values and it contains the true population parameter with in at certain confidence level of 90% , 95% , 99%. Given values are, Sample size n = 40 Sample standard deviation s = 4.1 Sample mean x̄ = 54 hours Formula : The confidence interval for true population mean, using t – distribution (since population standard deviation is not known) defined as follows Here, tα/2 is the t critical value with n-1 degrees of freedom at α% level of significance. Find the critical value tα/2: From the t – table values, Upper critical value tα/2 at 90% confidence level with 40-1 = 39 df is 1.685. i.e., tα/2, df = t0.1/2, 39 = 1.685 Therefore, the critical value tα/2 is 1.685. Substitute the values x̄ = 54 hours, s = 4.1, tα/2 = 1.685 & n = 40 in the above C.I formula Therefore, the 90% confidence interval estimate for the mean content viewing time per week is (52.9 to 55.1). Given values are, n = 40 x = 8 Sample proportion is p = x/n = 8/40 = 0.2 Formula : Formula of the confidence interval for the population proportion is given by where zα/2 is the critical value at α level of significance or at (1- α) % level of confidence. Critical value: zα/2 is the critical value at 90% confidence level is 1.645[ refer two tailed standard normal table values] Thus, zα/2 = z0.1/2 = 1.645 Submit the values p = 0.2 , zα/2 = 1.645 and n = 40 in the C.I formula as given Therefore, the 90% confidence interval for the true population proportion of residents who have HD on at least one television set is (0.096, 0.304) Given values are, confidence level = 90% population Standard deviation σ = 5 hours Margin of error M.E = 1 Formula : The formula for the Calculation of sample size is as following where zα/2 is the critical value at α level of significance or at (1- α) % level of confidence. Critical value: zα/2 is the critical value at 90% confidence level is 1.645[ refer two tailed standard normal table values] Thus, zα/2 = z0.1/2 = 1.645 Substitute the values zα/2 = 1.645, σ = 5 and M.E = 1 in the above sample size formula as given Hence, the required respondents are 68