ACTIVITY H: Sampling Theory and Estimation of Parameters 1. A random sample of 50 students in Holy Angel University shows that they spend an average of 3 hours per day on social media apps with a standard deviation of 0.5 hours. Assume a normal distribution. Construct a 90% and 97% confidence interval for the average number of hours spent on social media apps per day. 2. A study was conducted in which two types of fried chicken, Jollibee and KFC, were compared. Number of fried chickens sold, was measured. 90 Jollibee branches were surveyed while 80 were surveyed for KFC. The average fried chicken sold was 7,020 for Jollibee and 8,100 fried chickens for KFC. Find a 94% and 98% confidence interval on .Assume that the population standard deviations are 1000 and 500 for KFC and ????--?????????Jollibee, respectively. 3. The following are the average lengths of today’s top 10 Global Spotify hits, in minutes. 3.02, 2.85, 3.33, 3.43, 2.9, 2.93, 3.23, 2.77, 2.45, and 2.88. Find a 95% confidence interval for the variance of the lengths of this generation’s music, assuming a normal population.

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ACTIVITY H: Sampling Theory and Estimation of Parameters  1. A random sample of 50 students in Holy Angel University shows that they spend an average of 3 hours per day on social media apps with a standard deviation of 0.5 hours. Assume a normal distribution. Construct a 90% and 97% confidence interval for the average number of hours spent on social media apps per day.  2. A study was conducted in which two types of fried chicken, Jollibee and KFC, were compared. Number of fried chickens sold, was measured. 90 Jollibee branches were surveyed while 80 were surveyed for KFC. The average fried chicken sold was 7,020 for Jollibee and 8,100 fried chickens for KFC. Find a 94% and 98% confidence interval on .Assume that the population standard deviations are 1000 and 500 for KFC and ????--?????????Jollibee, respectively.  3. The following are the average lengths of today’s top 10 Global Spotify hits, in minutes. 3.02, 2.85, 3.33, 3.43, 2.9, 2.93, 3.23, 2.77, 2.45, and 2.88. Find a 95% confidence interval for the variance of the lengths of this generation’s music, assuming a normal population.

un4t5o4v · Accepted Answer

Part 1 As per bartleby guidelines only 1 question is answered, please upload the rest separately. The mean and standard deviation are μ=3,σ=0.5 respectively. For 90% confidence, the one tail test is 5% from right end that means it is 95% to right and 5% to the left. P(z≤x−μσ)=0.95 z=1.65 Hence the value of x is given by x=μ±zσ x=3±1.65×0.5 x=3.825,2.175 Therefore the interval is 2.175≤x≤3.825 for 90% confidence. Part 2 For 97% confidence, the one tail test is 1.5% from right end that means it is 98.5% to right and 1.5% to the left. P(z≤x−μσ)=0.985 z=2.17 Hence the value of x is given by x=μ±zσ x=3±2.17×0.5 x=4.085m 1.915 Therefore the interval is 1.915≤x≤4.085 for 97% confidence.

ACTIVITY H: Sampling Theory and Estimation of Parameters

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