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valentina_108 [34]
3 years ago
13

An online store randomly samples three groups of 100 customers on which delivery method they use. The data shows the results of

the three samples.
Overnight 2-Day Standard Total
Group 1 27 20 53 100
Group 2 32 17 51 100
Group 3 34 19 47 100



The online store has 2755 customers one day.

Based on the samples, what is a reasonable estimate of the number of people who will use overnight delivery?

Enter your answer rounded to the nearest whole number in the box.
Mathematics
2 answers:
masha68 [24]3 years ago
6 0

The answer is 854 for sure. I did the test and got it all right.

Sergio [31]3 years ago
3 0

Answer:

854 customers.

Step-by-step explanation:

An online store randomly samples three groups of 100 customers on which delivery method they use. The data shows the results of the three samples.

                          Overnight           2-day           Standard             Total

Group (1)                 27                     20                   53                   100

Group (2)                32                     17                    51                    100

Group (3)                34                      19                   47                    100

Total customers     93                     56                   151                  300

We have to calculate the number of people who will use overnight delivery.

Out of 300 customer there are 93 people who would use overnight delivery, so the percentage = \frac{93}{300} × 100 = 31%

The online store has 2755 customers one day.

Therefore, 31% of the 2755 customers = \frac{31}{2755} × 100 = 854.05 ≈ 854 customers.

854 customers would use overnight delivery.

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Chi square Daphne likes to ski at a resort that is open from December through April. According to a sign at the resort, 20, perc
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Answer:

There is not enough evidence to suggest that the snow falls in Daphne's hometown does not follow the given distribution.

Step-by-step explanation:

The missing data for the random sample of 80 days between December and April with snowfall is:

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December         16

January             11

February            16

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The Chi-square goodness of fit test would be used to determine whether the snow falls in Daphne's hometown followed the given distribution.

The hypothesis for the test can be defined as follows:

<em>H</em>₀: The snow falls in Daphne's hometown does not follow the given distribution.

<em>Hₐ</em>: The snow falls in Daphne's hometown followed the given distribution.

Assume that the significance level of the test is, <em>α</em> = 0.05.

The Chi-square test statistic is given by:

\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}

The value of Chi-square test statistic is computed in the Excel sheet below.

\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}=8.383

Compute the <em>p</em>-value as follows:

\text{p-value}=P(\chi^{2}_{(4)}>8.383)=CHISQ.DIST.RT(8.383,4)=0.0785

The <em>p</em>-value of the test is 0.0785.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

<em>p</em>-value = 0.0785 > <em>α</em> = 0.05.

The null hypothesis will not be rejected.

Conclusion:

There is not enough evidence to suggest that the snow falls in Daphne's hometown does not follow the given distribution.

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