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dlinn [17]
3 years ago
10

A newsletter publisher believes that 71q% of their readers own a personal computer. Is there sufficient evidence at the 0.010.01

level to refute the publisher's claim.
Required:
State the null and alternative hypotheses for the above scenario.
Mathematics
1 answer:
Natalka [10]3 years ago
4 0

Answer:

Null - p= 71%

Alternative - p =/ 71%

Step-by-step explanation:

The null hypothesis is always the default statement in an experiment. While the alternative hypothesis is always tested against the null hypothesis.

Null hypothesis: 71% of their readers own a personal computer- p = 71%

Alternative hypothesis: Not 71% of their readers own a personal computer - p =/ 71%

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chubhunter [2.5K]

Answer

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3 years ago
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Does the number 15/3 belong to the set of natural/counting numbers?
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No 15/3 does not belong to the set of numbers
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4 years ago
A local car dealer claims that 25% of all cars in San Francisco are blue. You take a random sample of 600 cars in San Francisco
sammy [17]

Answer:

No, we can't reject the dealer's claim with a significance level of 0.05.

Step-by-step explanation:

We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.

You take a random sample of 600 cars in San Francisco and find that 141 are blue.

<u><em>Let p = proportion of all cars in San Francisco who are blue</em></u>

SO, Null Hypothesis, H_0 : p = 25%   {means that 25% of all cars in San Francisco are blue}

Alternate Hypothesis, H_A : p \neq 25%   {means that % of all cars in San Francisco who are blue is different from 25%}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

                                  T.S.  = \frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }  ~ N(0,1)

where, \hat p = sample proportion of 600 cars in San Francisco who are blue =   \frac{141}{600} = 0.235

            n = sample of cars = 600

So, <u><em>test statistics</em></u>  =  \frac{0.235-0.25}{{\sqrt{\frac{0.235(1-0.235)}{600} } } } }

                               =  -0.866

<em>Now at 0.05 significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.</em>

Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.

4 0
3 years ago
You plan on financing a new road bike for $2,500. The bike shop offers a 13.5% APR for a 24 month loan. Use this information, an
Mila [183]
This question can be approached using the present value of annuity formula. The present value of annuity is given by PV=P\left( \frac{1-\left(1+ \frac{r}{t} \right)^{-nt}}{ \frac{r}{t} } \right), where: PV is the present value/amount of the loan, P is the periodic (monthly in this case) payment, r is the APR, t is the number of payments in one year and n is the number of years.

Given that the<span> financing is for a new road bike of $2,500 and that the bike shop offers a 13.5% APR for a 24 month loan.

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Therefore, his monthly payment is $119.44</span>
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h = 19 Meters

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8 0
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