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pickupchik [31]
4 years ago
10

On a typical day, a jewelry store sells 8 bracelets. The data show the number of bracelets sold on each day for four weeks at a

jewelry store.
Week 1: 7, 8, 7, 9, 1, 6, 8

Week 2: 6, 7, 8, 7, 9, 8, 17

Week 3: 3, 1, 2, 3, 2, 3, 2

Week 4: 9, 7, 8, 8, 11, 6, 7

For which week would the mean of the data be a good estimate of the jewelry store's weekly sales throughout the year?

CLEAR SUBMIT

y=9

y=3x

y=15

y=x+4
Mathematics
2 answers:
mote1985 [20]4 years ago
8 0

For which week would the mean of the data be a good estimate of the jewelry store's weekly sales throughout the year?

Answer: Week 4's mean would be a good estimate of the jewelry store's weekly sales throughout the year. Because the average of the Week 4's sales is 8. Also we can see the Week 4's data appears without outliers. So the option Week 4: 9, 7, 8, 8, 11, 6, 7  would be a good estimate of the jewelry store's weekly sales throughout the year.

<u>Explanation:</u>

Week 1's mean is:

\frac{7+8+7+9+1+6+8}{7} =\frac{46}{7} =6.57

Week 2's mean is:

\frac{6+7+8+7+9+8+17}{7} =\frac{62}{7} =8.86

Week 3's mean is:

\frac{3+1+2+3+2+3+2}{7} =\frac{16}{7} =2.29

Week 4's mean is:

 \frac{9+7+8+8+11+6+7}{7} =\frac{56}{7} =8

Since the week 4's mean is 8, which is same as typical day's sale of bracelet's. Therefore, week 4 will be good estimate of the jewelry store's weekly sales throughout the year.

Verdich [7]4 years ago
3 0

Answer:

week 4

Step-by-step explanation:

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A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Kar
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We conclude that the mean waiting time is less than 10 minutes.

Step-by-step explanation:

We are given that a public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes.

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Let \mu = <u><em>mean waiting time for bus number 14.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 10 minutes      {means that the mean waiting time is more than or equal to 10 minutes}

Alternate Hypothesis, H_A : \mu < 10 minutes    {means that the mean waiting time is less than 10 minutes}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                       T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

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So, <u><em>test statistics</em></u> =  \frac{7.8-10}{\frac{2.5}{\sqrt{18} } }  ~ t_1_7

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The value of t test statistics is -3.734.

Now, at 0.01 significance level the t table gives critical value of -2.567 for left-tailed test.

Since our test statistic is less than the critical value of t as -3.734 < -2.567, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the mean waiting time is less than 10 minutes.

5 0
4 years ago
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