4. The average annual income of 100 randomly chosen residents of Santa Cruz is $30,755 with a standard deviation of $20,450. a) What is the standard deviation of the annual income? b) Test the hypothesis that the average annual income is $32,000 against the alternative that it is less than $32,000 at the 10% level. c) Test the hypothesis that the average annual income is equal to $33,000 against the alternative that it is not at the 5% level. d) What is the 95% confidence interval of the average annual income?

a) The standard deviation of the annual income σₓ = 2045

b)

The calculated value Z = 0.608 < 1.645 at 10 % level of significance

Null hypothesis is accepted

The average annual income is greater than $32,000

c)

The calculated value Z = 1.0977 < 1.96 at 5 % level of significance

Null hypothesis is accepted

The average annual income is  equal to  $33,000

d)

95% of confidence intervals of the Average annual income

(26 ,746.8 ,34, 763.2)

Step-by-step explanation:

Given size of the sample 'n' =100

mean of the sample x⁻ =  $30,755

The Standard deviation = $20,450

a)

The standard deviation of the annual income σₓ =

                                               =

b)

Given mean of the Population μ =  $32,000

Given size of the sample 'n' =100

mean of the sample x⁻ =  $30,755

The Standard deviation ( σ)= $20,450

Null Hypothesis:- H₀: μ > $32,000

Alternative Hypothesis:H₁: μ <  $32,000

Level of significance α = 0.10

Z= |-0.608| = 0.608

The calculated value Z = 0.608 < 1.645 at 10 % level of significance

Null hypothesis is accepted

The average annual income is  greater than $32,000

c)

Given mean of the Population μ =  $33,000

Given size of the sample 'n' =100

mean of the sample x⁻ =  $30,755

The Standard deviation ( σ)= $20,450

Null Hypothesis:- H₀: μ =  $33,000

Alternative Hypothesis:H₁: μ ≠ $33,000

Level of significance α = 0.05

Z = -1.0977

|Z|= |-1.0977| = 1.0977

The 95% of z -value = 1.96

The calculated value Z = 1.0977 < 1.96 at 5 % level of significance

Null hypothesis is accepted

The average annual income is equal to  $33,000

d)

95% of confidence intervals is determined by

( 30 755 - 4008.2 , 30 755 +4008.2)

95% of confidence intervals of the Average annual income

(26 ,746.8 ,34, 763.2)

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!