Question

1) You measure 25 textbook' weights, and find they have a mean weight of 60 ounces. Assume the population standard deviation is 9.2 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean backpack weight.

Give your answer as a decimal, to two places

a) You measure 46 textbooks' weights, and find they have a mean weight of 48 ounces. Assume the population standard deviation is 7.8 ounces. Based on this, construct a 99% confidence interval for the true population mean textbook weight.

Give your answers as decimals, to two places

b) You measure 40 textbooks weights, and find they have a mean weight of 52 ounces. Assume the population standard deviation is 12.1 ounces. Based on this, construct a 99% confidence interval for the true population mean watermelon weight.

Give your answers as decimals, to two places

Answer 1

1)

Margin of error = Z/2 * / sqrt(n)

= 1.645 * 9.2 / sqrt(25)

= 3.03

2)

99% Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.01 /2) = 2.576
48 ± Z (0.01/2 ) * 7.8/√(46)
Lower Limit = 48 - Z(0.01/2) 7.8/√(46)
Lower Limit = 45.04
Upper Limit = 48 + Z(0.01/2) 7.8/√(46)
Upper Limit = 50.96
99% Confidence interval is ( 45.04 , 50.96 )

3)

99% Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.01 /2) = 2.576
52 ± Z (0.01/2 ) * 12.1/√(40)
Lower Limit = 52 - Z(0.01/2) 12.1/√(40)
Lower Limit = 47.07
Upper Limit = 52 + Z(0.01/2) 12.1/√(40)
Upper Limit = 56.93
99% Confidence interval is ( 47.07 , 56.93 )