You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the po

Tazmin Horton

Tazmin Horton

Answered question

2020-12-15

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. Interpret the results and compare the widths of the confidence intervals. From a random sample of 58 ​dates, the mean record high daily temperature in a certain city has a mean of 83.43F. Assume the population standard deviation is 14.02F. 90%= 95%= Which interval is wider?

Answer & Explanation

Obiajulu

Obiajulu

Skilled2020-12-16Added 98 answers

Step 1 Given Data: The sample size is: n=58 The mean temperature of the city is: μ=83.43F The standard deviation of the temperature is: σ=14.02F 95% Population: The expression calculate the 95% of the Population is, 12α=0.95
α=10.952
=0.025 Corresponds to probability 0.025, the z value from the right tail end of the z-table 1.96. Step 2 The expression to calculate the interval of the population is, CI=X±M.E
=X±Z×σn Substitute values in the above expression. CI=83.43±1.96×14.0258
=83.43±3.60
=(83.433.60,83.43+3.60)
(79.83,87.03) Thus, the interval width of the 95% level is 3.6. and the interval is (79.83, 87.03). Step 3 90% Population: The expression calculate the 95% of the Population is, 12α=0.90
α=10.902
=0.050 Corresponds to probability 0.050, the z value from the right tail end of the z-table is 1.64. The expression to calculate the interval of the Population is, CI=X±M.E
=X±Z×σn Substitute values in the above expression. CI=83.43±1.96×14.0258
=83.43±3.02
=(83.433.60,83.43+3.02)
=(80.41,86.45) Thus, the interval width of 90% population level is 3.02, and the interval is (80.41, 59.45). Thus, 95% population has higher interval width.

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