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anygoal [31]
3 years ago
11

A marketing firm is studying consumer preferences for winter fashions in four different months. From a population of women 18-21

years of age, a random sample of 100 women was selected in January. Another random sample of 100 women was selected in March. Another random sample of 100 women was selected in June. Another random sample of 100 women was selected in September. What is the number of samples?
Mathematics
1 answer:
Solnce55 [7]3 years ago
4 0

Answer:

There were total 4 samples collected and each sample had a sample size of n = 100.

Step-by-step explanation:

We are given the following information in the question:

Collection of samples:

1^{st}\text{ sample was done in January with a sample size n = 100}

2^{nd}\text{ sample was done in March with a sample size n = 100}

3^{rd}\text{ sample was done in June with a sample size n = 100}

4^{th}\text{ sample was done in September with a sample size n = 100}

Hence, there were total 4 samples collected and each sample had a sample size of n = 100.

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Answer:

3, 4, and 5

Step-by-step explanation:

To put it simply, 3 + 4 + 5 = 12 and the numbers are consecutive, next to each other.

The method:

The three consecutive numbers could be called x, x + 1, and x + 2 and we know that x + x + 1 + x + 2 = 12

This can be simplified to 3x + 3 = 12. Next, subtract 3 from both sides to get 3x = 9. Now, divide both sides by 3 to get x = 3.

This means that the first number is 3 and the other numbers are 4 (3 + 1) and 5 (3 + 2).

Hope this helps!

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<em>Answer below:</em>



<em>Hope this helps :-)</em>

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The valid conclusions for the manager based on the considered test is given by: Option

<h3>When do we perform one sample z-test?</h3>

One sample z-test is performed if the sample size is large enough (n  > 30) and we want to know if the sample comes from the specific population.

For this case, we're specified that:

  • Population mean = \mu = $150
  • Population standard deviation = \sigma = $30.20
  • Sample mean = \overline{x} = $160
  • Sample size = n = 40 > 30
  • Level of significance = \alpha = 2.5% = 0.025
  • We want to determine if the average customer spends more in his store than the national average.

Forming hypotheses:

  • Null Hypothesis: Nullifies what we're trying to determine. Assumes that the average customer doesn't spend more in the store than the national average. Symbolically, we get: H_0: \mu_0 \leq \mu = 150
  • Alternate hypothesis: Assumes that customer spends more in his store than the national average. Symbolically H_1: \mu_0 > \mu = 150

where \mu_0 is the hypothesized population mean of the money his customer spends in his store.

The z-test statistic we get is:

z = \dfrac{\overline{x} - \mu_0}{\sigma/\sqrt{n}} = \dfrac{160 - 150}{30.20/\sqrt{40}} \approx 2.094

The test is single tailed, (right tailed).

The critical value of z at level of significance 0.025 is 1.96

Since we've got 2.904 > 1.96, so we reject the null hypothesis.

(as for right tailed test, we reject null hypothesis if the test statistic is > critical value).

Thus, we accept the alternate hypothesis that customer spends more in his store than the national average.

Learn more about one-sample z-test here:

brainly.com/question/21477856

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