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NeX [460]
4 years ago
11

Do consumers spend more on a trip to Target or​ Walmart? Suppose researchers interested in this question collected a systematic

sample for 84 Target customers and 80 Walmart customers by asking customers for their purchase amount as they left the store. Using the given summary​ statistics, researchers calculated a​ 95% confidence interval for the mean difference between Target and Walmart purchase amounts. The interval was ​($negative 15.05​,$negative 2.95​). Explain in context what this interval means.
Mathematics
1 answer:
ra1l [238]4 years ago
5 0

Answer:

Step-by-step explanation:

Confidence interval for the difference between two population means is written in the form,

difference in sample means ± margin of error

The difference in sample means is the point estimate for the difference in population means. In the given scenario, the point estimate is the difference in mean amount spent by the sampled customers on a trip to Target or​ Walmart.

Since the interval was (- $15.05​,$2.95), it means that the lower limit is - $15.05 and the upper limit us $2.95.

Therefore, the 95% confidence interval is providing a range that we are 95% confident that the true difference in mean amount spent by Target customers and Walmart customers falls between - $15.05​ and $2.95

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Help please, will mark brainliest ​
mario62 [17]

Answer:

Step-by-step explanation:

LET ORANGES=X

PEARS ARE 3 TIMES =3X

LET CHILDREN=Y

IF WE GIVE 5 ORANGES PER CHILD WE NEED 5Y ORANGES.SO THIS =X

X=5Y........................I

IF WE GIVE 8 PEARS PER CHILD WE NEED 8Y PERARS...THERE ARE STILL 21 PEARS LEFT OUT AFTER THIS THAT IS

8Y+21=3X....................II

SUBSTITUTING FOR X FROM EQN.I..

8Y+21=3*5Y

15Y-8Y=21

7Y=21

Y=3

X==5Y=5*3=15..

SO CHILDREN =3

ORANGES=15

PEARS=3*15=45

THANKYOU AND PLEASE AMRK IT BRAINLIEST

8 0
3 years ago
It’s a yes or no question<br> !!!!!!!!!!!!<br> For math
vesna_86 [32]
I’m confused on what the question is exactly
5 0
3 years ago
Find the quotient. Round to the nearest tenth. 3784 divided by 18
SCORPION-xisa [38]

Answer:

210.22

Step-by-step explanation:

first you need to divide 3785 divided by 18 which is 210.22222222 then u need to round to the nearest tenth which is equaled to 210.22. I hope this helped

5 0
3 years ago
Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology ex
Nataly [62]

Answer:

(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942  

(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978  

And the 90% confidence interval would be given (0.0942;0.09978).  

We are confident at 90% that the difference between the two proportions is between 0.0942 \leq p_A -p_B \leq 0.09978

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

p_A represent the real population proportion female for Biology

\hat p_A =\frac{84199}{144796}=0.582 represent the estimated proportion female for biology

n_A=144796 is the sample size for A

p_B represent the real population proportion female for calculus AB

\hat p_B =\frac{102598}{211693}=0.485 represent the estimated proportion female for Calculus AB

n_B=211693 is the sample size required for B

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 90% confidence interval the value of \alpha=1-0.90=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.64  

And replacing into the confidence interval formula we got:  

(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942  

(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978  

And the 90% confidence interval would be given (0.0942;0.09978).  

We are confident at 90% that the difference between the two proportions is between 0.0942 \leq p_A -p_B \leq 0.09978

5 0
3 years ago
Find the variance of this probability distribution. Round to two decimal places.​
attashe74 [19]

Answer:

Variance = 4.68

Step-by-step explanation:

The formula for the variance is:

\sigma^{2} =\frac{\Sigma(X- \mu)^{2}}{N} \\or \\ \sigma^{2} =\frac{\Sigma(X)^{2}}{N} -\mu^{2} \\

Where:

X: Values \\\mu: Mean \\N: Number\ of\ values

The mean can be calculated as each value multiplied by its probability

\mu = 0*0.4 + 1*0.3 + 2*0.1+3*0.15+ 4*0.05=1.15

\frac{\Sigma (X)^{2}}{N} =\frac{(0^{2}+1^{2}+2^{2}+3^{2}+4^{2})}{5} =6

Replacing the mean and the summatory of X:

\sigma^{2} = \frac{\Sigma(X)^{2}}{N} -\mu^{2} \\= 6 - 1.15^{2}\\= 4.6775

4 0
3 years ago
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