Question

Based on past data, the sample mean of the credit card purchases at a large department store is $45. Assuming sample size is 25 and the sample standard deviation is 9.

a)          What % of samples are likely to have between 25 and 35?                                                                  ___________

b)         Between what two values 95% of sample means fall?                                                                         ___________

c)          Below what value 98% of sample means fall?                                                                                   ___________

d)         Above what value only5% of sample means fall??                                                                             ___________

Within what symmetrical limits of the population percentage will 95% of the sample percentages fall?

Answer 1

SOLUTION:

From given data,

Based on past data, the sample mean of the credit card purchases at a large department store is $45. Assuming sample size is 25 and the sample standard deviation is 9.

Sample size = n = 25

Standard deviation = = 9

Mean = = 45   

z = (x - ) / = (x - 45) / 9

a) What % of samples are likely to have between 25 and 35?

z value for 25,

z = (25-45)/9 = -2.22

z value for 35,

z = (35-45)/9 = -1.11

P( 25 < x < 35) = P( -2.22 < z < -1.11)

P( 25 < x < 35) = P( z < -1.11) – P( Z< -2.22)

P( 25 < x < 35) = 0.13350 - 0.01321

P( 25 < x < 35) = 0.12029

b) Between what two values 95% of sample means fall?

95% confidence interval

Confidence interval is 95%

95% = 95/100 = 0.95

= 1 - Confidence interval = 1-0.95 = 0.05

/2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Standard error = / sqrt(n)
Standard error  =9 /sqrt(25)

Standard error = SE =1.8

Confidence interval is 95% = Z_/2 * SE

Lower value = - Z_/2 * / sqrt(n) =45 - 1.96 * 1.8 = 41.472

upper value = + Z_/2 * / sqrt(n) =45 + 1.96 * 1.8 = 48.528

c) Below what value 98% of sample means fall?

98% confidence interval

Confidence interval is 98%

98% = 98/100 = 0.98

= 1 - Confidence interval = 1-0.98 =0.02

/2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.3263

The required value is

=    + Z_/2 * / sqrt(n)

= 45 +2.3263 * 1.8

= 49.18734

d ) Above what value only 5% of sample means fall?

5% = 5/100 = 0.05

z_critical = 1.6449

The required value

= ​​​​​​​ + Z_/2 * / sqrt(n)

=45 + 1.6449* 1.8

= 47.96082

Within what symmetrical limits of the population percentage will 95% of the sample percentages fall?

95% confidence interval

Confidence interval is 95%

95% = 95/100 = 0.95

= 1 - Confidence interval = 1-0.95 = 0.05

/2 = 0.05 / 2

= 0.025

Z_/2 = Z_0.025 = 1.96

Standard error = / sqrt(n)
Standard error  =9 /sqrt(25)

Standard error = SE =1.8

Confidence interval is 95% = Z_/2 * SE

Lower value = - Z_/2 * / sqrt(n) =45 - 1.96 * 1.8 = 41.472

upper value = + Z_/2 * / sqrt(n) =45 + 1.96 * 1.8 = 48.528