Consider the following two questions designed to assess quantitative literacy: a. What is 15% of 1000? b. A store is offering a 1596 off sale on all TVs. The most popular television is normally priced at $1000. How much money would a customer save on the television during this sale? Suppose the first question is asked of 200 randomly selected college students, with 165 answering correctly; the second one is asked of a different random sample of 200 college students, resulting in 142 correct responses. Carry out a test of hypotheses at significance level 0.05 to decide if the true proportion of correct responses to the question without context exceeds that for the one with context. (Use p1 for the true proportion students who answered the question without context correctly and p2 for the true proportion of students who answered the question with context correctly.) Carry out a test of hypotheses at significance level 0.05 to decide if the true proportion of correct responses to the question without context exceeds that for the one with context.

The calculated value of z= 2.7225 falls in the critical region therefore we reject the null hypothesis and conclude that at the  5% significance level, true proportion of correct responses to the question without context exceeds that for the one with context.

Step-by-step explanation:

a: 15 % of 1000= 150

b. 15.96 % of $ 1000= $ 159.6

The customer would pay = $ 1000- $ 159.6= $ 840.4 and save $ 159.6

Formulate the hypotheses as

H0: p1= p2   true proportion of correct responses to the question without context is equal that for the one with context.

Ha : p1≠ p2

We choose the significance level ∝= 0.05

The critical value for two tailed test at alpha=0.05 is ± 1.96

The test statistic is

Z = p1-p2/ √pq (1/n1+ 1/n2)

p1= true proportion students who answered the question without context correctly  = 165/200=0.825

p2=  true proportion of students who answered the question with context correctly = 142/200= 0.71

p = an estimate of the common rate on the assumption that the two proportions are same.

p = n1p1+ n2p2/ n1 + n2

p =200 (0.825) + 200 (0.71) / 400

p=  165+ 142/400= 307 /400 =0.7675

now q = 1-p= 1- 0.7675= 0.2325

Thus

z= 0.825- 0.71/ √0.7675*0.2325( 1/200 + 1/200)

z= 0.115/√ 0.17844( 2/200)

z= 0.115/0.04224

z= 2.7225

The calculated value of z falls in the critical region therefore we reject the null hypothesis and conclude that at the  5% significance level, true proportion of correct responses to the question without context exceeds that for the one with context.

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!

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step-by-step explanation: