Mathematics, 26.03.2020 02:06 chickenndle7814

In 1990, 23.8% of 18-24 year-olds in Biddeford, Maine were attending college. About 10 years later, a random sample of 610 18-24 year-olds showed 178 were attending college. At 1%

significance, does this indicate that the percentage of residents attending college in that age-group is greater than 23.8%?

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Answers

There is enough evidence to support the claim that the percentage of residents attending college in that age-group is greater than 23.8%

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 610

p = 23.8% = 0.238

Alpha, α = 0.01

Number of 18-24 year-old attending college , x = 178

First, we design the null and the alternate hypothesis

$H_{0}: p = 0.238\\H_A: p 0.238$

This is a one-tailed(right) test.

Formula:

$\hat{p} = \dfrac{x}{n} = \dfrac{178}{610} = 0.2918$

$z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Putting the values, we get,

$z = \displaystyle\frac{0.2918-0.238}{\sqrt{\frac{0.238(1-0.238)}{610}}} = 3.1201$

Now, we calculate the p-value from the table.

P-value = 0.0009

Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject the null hypothesis.

Conclusion:

Thus there is enough evidence to support the claim that the percentage of residents attending college in that age-group is greater than 23.8%

Answer from: isabellainksow87vn

We conclude that the percentage of residents attending college in that age-group is greater than 23.8%.

Step-by-step explanation:

We are given that in 1990, 23.8% of 18-24 year-old in Biddeford, Maine were attending college. About 10 years later, a random sample of 610 18-24 year-old showed 178 were attending college.

Let p = % of 18-24 year-old in Biddeford, Maine who were attending college = 23.8%

Let Null Hypothesis, H_0 : p $\leq$ 0.238 {means that the percentage of residents attending college in that age-group is less than or equal to 23.8%}

Alternate Hypothesis, H_a : p > 0.238 {means that the percentage of residents attending college in that age-group is greater than 23.8%}

The test statistics will be used here is One-sample proportion test;

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

where, $\hat p$ = % of 18-24 year-old in Biddeford, Maine who were attending college in a sample of 610 people = $\frac{178}{610}$ = 0.292

n = sample size = 610

So, test statistics = $\frac{0.292-0.238}{\sqrt{\frac{0.292(1-0.292)}{610} } }$

= 2.933

Now, at 1% significance level the z table gives critical value of 2.5758. Since our test statistics is more than the critical value so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the percentage of residents attending college in that age-group is greater than 23.8%.

Answer from: Quest

d is the correct answer

step-by-step explanation:

x= 8 because 3x = 3 * 8, and 3*8 = 24

Answer from: Quest

so, converting firstly the mixed fractions to improper fractions and then "adding".

$\bf \stackrel{mixed}{2\frac{9}{10}}\implies \cfrac{2\cdot 10+9}{10}\implies \stackrel{improper}{\cfrac{29}{10}}~\hfill \stackrel{mixed}{5\frac{4}{15}}\implies \cfrac{5\cdot 15+4}{15}\implies \stackrel{improper}{\cfrac{79}{15}} [-0.35em] ~\dotfill$

$\bf -\cfrac{29}{10}+\left( -\cfrac{79}{15} \right)\implies -\cfrac{29}{10}-\cfrac{79}{15}\implies \stackrel{\textit{using the lcd of 30}}{\cfrac{-(3)29-(2)79}{30}}\implies \cfrac{-87-158}{30} \cfrac{-245}{30}\implies -\cfrac{49}{6}\implies -8\frac{1}{6}$