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2 years ago

A man sells a sofa for rs 3210 making a profit of 7% what would have been his profit percent if he had sold it for rs 3360?

1 answer:

7 0

Let x be the original cost for the sofa.
X x (1 + 0.07) =3210
X = 3000
Now we know the cost for the sofa.
Let p be the profit percent.
3000 x (1 + p) = 3360
P = 0.12
So the percentage profit is 12%

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2 years ago

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1 year ago

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2 years ago

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Last year, 46% of business owners gave a holiday gift to their employees. A survey of business owners indicated that 35% plan to

DiKsa [7]

Answer:

a) Number = 60 *0.35=21

b) Since is a left tailed test the p value would be:

$p_v =P(Z$

c) If we compare the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.

We can use as smallest significance level 0.044 and we got the same conclusion.

Step-by-step explanation:

Data given and notation

n=60 represent the random sample taken

X represent the business owners plan to provide a holiday gift to their employees

$\hat p=0.35$ estimated proportion of business owners plan to provide a holiday gift to their employees

$p_o=0.46$ 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 (variable of interest)

$p_v$ represent the p value (variable of interest)

Part a

On this case w ejust need to multiply the value of th sample size by the proportion given like this:

Number = 60 *0.35=21

2) Part b

We need to conduct a hypothesis in order to test the claim that the proportion of business owners providing holiday gifts had decreased from the 2008 level:

Null hypothesis: $p\geq 0.46$

Alternative hypothesis: $p < 0.46$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.35 -0.46}{\sqrt{\frac{0.46(1-0.46)}{60}}}=-1.710$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(Z$

Part c

If we compare the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.

We can use as smallest significance level 0.044 and we got the same conclusion.

8 0

2 years ago

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

andrezito [222]

Answer:

a) Null hypothesis: $p \leq 0.113$

Alternative hypothesis: $p > 0.113$

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

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

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%

8 0

2 years ago