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Scrat [10]
3 years ago
6

The football team has 50 jerseys. There are 15 medium-sized jerseys. What percent of the jerseys are medium sized jerseys?

Mathematics
1 answer:
icang [17]3 years ago
8 0
15/50 = 3/30 = 1/10 1/10 = 10% That is about as simple as I could put it.
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<h2>x = 129</h2>

Step-by-step explanation:

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The hypothesis test shows that we reject the null hypothesis and there is sufficient evidence to support the claim that the return rate is less than 20%

<h3>What is the claim that the return rate is less than 20% by using a statistical hypothesis method?</h3>

The claim that the return rate is less than 20% is p < 0.2. From the given information, we can compute our null hypothesis and alternative hypothesis as:

\mathbf{H_o :p =0.2}

\mathbf{H_i:p < 0.2}

Given that:
Sample size (n) = 6965

Sample proportion \mathbf{\hat p = \dfrac{x}{n} = \dfrac{1302}{6965} \sim0.1869}

The test statistics for this data can be computed as:

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

\mathbf{z = \dfrac{0.1869 -0.2}{\sqrt{\dfrac{0.2(1-0.2)}{6965}}}}

\mathbf{z = \dfrac{-0.0131}{0.0047929}}

z = -2.73

From the hypothesis testing, since the p < alternative hypothesis, then our test is a left-tailed test(one-tailed.

Hence, the p-value for the test statistics can be computed as:

P-value = P(Z ≤ z)

P-value = P(Z ≤ - 2.73)

By using the Excel function =NORMDIST (-2.73)

P-value = 0.00317

P-value ≅ 0.003

Therefore, we can conclude that since P-value is less than the significance level at ∝ = 0.01, we reject the null hypothesis and there is sufficient evidence to support the claim that the return rate is less than 20%

Learn more about hypothesis testing here:

brainly.com/question/15980493

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3 0
1 year ago
A new sample of 225 employed adults is chosen. Find the probability that less than 7.1% of the individuals in this sample hold m
arsen [322]

Answer:

The probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

Step-by-step explanation:

Let <em>X</em> = number of individuals in the United States who held multiple jobs.

The probability that an individual holds multiple jobs is, <em>p</em> = 0.13.

The sample of employed individuals selected is of size, <em>n</em> = 225.

An individual holding multiple jobs is independent of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 225 and <em>p</em> = 0.13.

But since the sample size is too large Normal approximation to Binomial can be used to define the distribution of proportion <em>p</em>.

Conditions of Normal approximation to Binomial are:

  • np ≥ 10
  • n (1 - p) ≥ 10

Check the conditions as follows:

np=225\times 0.13=29.25>10\\n(1-p)=225\times (1-0.13)=195.75>10

The distribution of the proportion of individuals who hold multiple jobs is,

p\sim N(p, \frac{p(1-p)}{n})

Compute the probability that less than 7.1% of the individuals in this sample hold multiple jobs as follows:

P(p

*Use a <em>z</em>-table.

Thus, the probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

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Divison probem that equals 36
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36 Divided by 6 :]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]
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