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2 years ago

In general, if the sample size is large, adding an extremely large outlier to a data set will not substantially affect _____ .

1 answer:

4 0

In general, if the sample size is large, adding an extremely large outlier to a data set will not substantially affect _________

a. the sample standard deviation

b. the sample mean

c. the sample median

d. none of the above.

Answer: We know that the median is least affected by outliers, because adding an outlier will only affect the tails not the center of the data distribution. Therefore, adding an extremely large outlier to a data set will not substantially affect the sample median. Hence the option C. the sample median is correct

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Joseph claims that a scatterplot in which the y-values increase as the x-values increase must have a linear association. Amy cla

quester [9]

Answer:

Amy is correct because a nonlinear association could increase along the whole data set, while being steeper in some parts than others. The scatterplot could be linear or nonlinear.

Step-by-step explanation:

I just took the quiz, I hope this helps :).

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2 years ago

Read 2 more answers

The equation y = 5.5x represents a proportional relationship. What is the constant of proportionality?

lara31 [8.8K]

Y = 5.5x
The constant is 5.5
since no matter what y and x are there will always be a 5.5 difference

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2 years ago

Carol is going to buy a purse she found. It is originally $120.00, but has been marked down 25%. When Carol goes to buy the purs

Reil [10]

$82.80
I hope this helped!

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2 years ago

Direct Mailing Company sells computers and computer parts by mail. The company claims that more than 90% of all orders are maile

lys-0071 [83]

Answer:

We conclude that less than or equal to 90% of all orders are mailed within 72 hours after they are received which means the company's claim is not true.

Step-by-step explanation:

We are given that the company claims that more than 90% of all orders are mailed within 72 hours after they are received. The quality control department at the company often takes samples to check if this claim is valid.

A recently taken sample of 175 orders showed that 161 of them were mailed within 72 hours.

Let p = percentage of all orders that are mailed within 72 hours.

SO, Null Hypothesis, $H_0$ : p $\leq$ 90% {means that less than or equal to 90% of all orders are mailed within 72 hours after they are received}

Alternate Hypothesis, $H_A$ : p > 90% {means that more than 90% of all orders are mailed within 72 hours after they are received}

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

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

where, $\hat p$ = proportion of orders that were mailed within 72 hours in a sample of 175 = $\frac{161}{175}\times 100$ = 92%

n = sample of orders = 175

So, test statistics = $\frac{0.92-0.90}{{\sqrt{\frac{0.92(1-0.92)}{175} } } } }$

= 0.975

The value of the test statistics is 0.975.

Now at 1% significance level, the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is less than the critical value of z as 0.975 < 2.3263, 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.

Therefore, we conclude that less than or equal to 90% of all orders are mailed within 72 hours after they are received which means the company's claim is not true.

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2 years ago

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

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

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.

So, 90% confidence interval for the population proportion, p is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$