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1 year ago

8

A mattress store sold 1,330 mattresses in one year. If 30% of the mattresses sold were king size mattresses, how many king size

mattresses did the store sell

2 answers:

Ksju [112]1 year ago

4 0

Answer: 399

Step-by-step explanation:

Given : A mattress store sold 1,330 mattresses in one year.

If 30% of the mattresses sold were king size mattresses , then the number of king size mattresses would be

30% of 1330

Replace 'of' be Multiply sign '×' and convert percent into decimal by divide it by 100 , we get

$\dfrac{30}{100}\times1330=\dfrac{39900}{100}=399$

Therefore , the number of king size mattresses were sold = 399

kodGreya [7K]1 year ago

3 0

The total number of mattresses that are sold in a year = 1,330

Percentage of King size mattresses sold = 30% = 30/100 = 0.3

Next, we have to calculate the actual number sold by using the formula given below:

Number of king size mattresses sold = Total mattresses sold in a year * percentage of king size mattresses sold

Number of King Size mattresses sold = 1,330*0.3 = 399

Number of king size mattresses sold = 399

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Please find the attachment for visual understanding.

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The U.S. Bureau of Labor Statistics reports that 11.3% of U.S. workers belong to unions (BLS website, January 2014). Suppose a s

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a) Null hypothesis: $p \leq 0.113$

Alternative hypothesis: $p > 0.113$

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Replacing we got:

$z=\frac{0.13 -0.113}{\sqrt{\frac{0.113(1-0.113)}{400}}}=1.074$

The p value for this case would be given by:

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c) For this case we see that the p value is higher than the significance level of 0.05 so then we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion workers belonged to unions is significantly higher than 11.3%

Step-by-step explanation:

Information given

n=400 represent the random sample taken

X=52 represent the workers belonged to unions

$\hat p=\frac{52}{400}=0.13$ estimated proportion of workers belonged to unions

$p_o=0.113$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic

$p_v$ represent the p value

Part a

We want to test if the true proportion of interest is higher than 0.113 so then the system of hypothesis are.:

Null hypothesis: $p \leq 0.113$

Alternative hypothesis: $p > 0.113$

Part b

The statistic is given by:

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

Replacing we got:

$z=\frac{0.13 -0.113}{\sqrt{\frac{0.113(1-0.113)}{400}}}=1.074$

The p value for this case would be given by:

$p_v =P(z>1.074)=0.141$

Part c

For this case we see that the p value is higher than the significance level of 0.05 so then we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion workers belonged to unions is significantly higher than 11.3%

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