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Blizzard [7]
3 years ago
7

Maylin goes to her local sporting goods store to buy socks and jogging shorts. Socks cost​ $10 per pair and jogging shorts cost​

$15 each. At her current consumption bundle pair of 4 pairs of socks and 2 jogging​ shorts, her marginal benefit from the last pair of socks is 20 and the marginal benefit of the last jogging short is 30. Maylin should​ ________.
Mathematics
1 answer:
PolarNik [594]3 years ago
5 0

Answer:

Maylin should purchase one unit of each socks & jogging shorts.

Step-by-step explanation:

In case of two goods, consumer is at equilibrium when marginal utility per dollar spent on last unit of the goods are equal, ie MU 1 / P1 = MU 2 / P2 (MU are the marginal utilities / marginal benefit of two goods, P are the price of two goods)

Given : MU1 (So) = 20, MU2 (JSh)  = 30, P1 (So) = 10, P2 (JSh) = 15

So, equilibrium is at where MU1 (So) / P1 (So) = MU2 (JSh) / P2 (Jsh)

20/10 = 2, & 30/15 = 2.  MU's per dollar are equal for both.

So Maylin is in equilibrium, she should purchase one unit of each socks & jogging shorts.

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A Pew Internet poll asked cell phone owners about how they used their cell phones. One question asked whether or not during the
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Answer:

a) \hat p=\frac{471}{1024}=0.460

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SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156

And the margin of error is given by:

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b) The 99% confidence interval would be given by (0.429;0.491)

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

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Data given and notation  

n=1024 represent the random sample taken    

X=471 represent the people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

\hat p=\frac{471}{1024}=0.460 estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering    

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Part a

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The standard error is given by:

SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156

And the margin of error is given by:

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Part b

If we replace the values obtained we got:

0.460-1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.429

0.460+1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.491

The 99% confidence interval would be given by (0.429;0.491)

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