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Brums [2.3K]
3 years ago
14

A distributor of personal computers has five locations in a city. In the year's first quarter, the sales in units were: Location

Observed Sale ( units) Northside 70 Pleasant Township 75 Southwick 70 I-90 50 Venice Avenue 35 Total 300 For a goodness-of-fit test that sales were the same for all locations, what is the critical value at the 0.01 level of risk? Select one: a. 7.779 b. 15.033 c. 13.277 d. 5.412

Mathematics
1 answer:
Deffense [45]3 years ago
7 0

Answer:

The correct option is (c) 13.277.

Step-by-step explanation:

The observed data is:

Location                      Observed sales (units)

Northside                                70

Pleasant Township                 75

Southwick                               70

I-90                                          50

Venice Avenue                       35

TOTAL                                   300

The test statistic for the Goodness of fit test is:

\chi^{2}=\sum\frac{(Observed-Expected)^{2}}{Expected}

For <em>k</em> independent samples this test statistic follows a Chi-square distribution with degrees of freedom (<em>k</em> - 1).

The sample size in this case is, <em>k</em> = 5.

The degrees of freedom is, (<em>k</em> - 1) = 4.

The level of significance is, <em>α</em> = 0.01.

The critical value of the test is:

\chi^{2}_{\alpha ,k-1}=\chi^{2}_{0.01,4}=13.277

**Use the Chi-square table.

Thus, the critical value is 13.277.

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A random sample of 20 purchases showed the amounts in the table (in $). The mean is $49.57 and the standard deviation is $20.28.
son4ous [18]

Answer:

The 98​% confidence interval for the mean purchases of all​ customers is ($37.40, $61.74).

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.98}{2} = 0.01

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.01 = 0.99, so z = 2.325

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 2.325*\frac{20.28}{\sqrt{20}} = 12.17

The lower end of the interval is the mean subtracted by M. So it is 49.57 - 12.17 = $37.40.

The upper end of the interval is the mean added to M. So it is 49.57 + 12.17 = $61.74.

The 98​% confidence interval for the mean purchases of all​ customers is ($37.40, $61.74).

3 0
3 years ago
A flock of griffins are flying together along the same route as the dragons. Griffins have the head and wings of an eagle, but t
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3 years ago
Solve : 6 yards 9 feet 4 inches times 2
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Help, i know the answer but i don’t know how to do the work
Diano4ka-milaya [45]

Answer:

C

Step-by-step explanation:

Using the rule of radicals

\sqrt{a} × \sqrt{b} ⇔ \sqrt{ab}

Simplifying each radical before combining them.

\sqrt{45t}

= \sqrt{9(5t)}

= \sqrt{9} × \sqrt{5t} = 3\sqrt{5t}

-----------------------------------------------------------------------------

\sqrt{125t}

= \sqrt{25(5t)}

= \sqrt{25} × \sqrt{5t} = 5\sqrt{5t}

---------------------------------------------------------------------------

Hence

2(3\sqrt{5t}) - 3(5\sqrt{5t}

= 6\sqrt{5t} - 15\sqrt{5t}

= - 9\sqrt{5t} → C

7 0
4 years ago
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