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mojhsa [17]
1 year ago
10

Market research was undertaken to determine whether an ad campaign was successful for the software company. 15 men and 75 women

were asked if they had seen the company’s latest TV advertisement. Altogether, 2 5 of the people said ‘yes’. 6 15 of the men said ‘yes’. What percentage of women said ‘no’ out of the total survey population?
Mathematics
1 answer:
rosijanka [135]1 year ago
4 0

50% of women who said ‘no’ out of the total survey population if 15 men and 75 women were asked if they had seen the company’s latest TV advertisement.

<h3>What is the percentage?</h3>

It's the ratio of two integers stated as a fraction of a hundred parts. It is a metric for comparing two sets of data, and it is expressed as a percentage using the percent symbol.

It is given that:

Market research was undertaken to determine whether an ad campaign was successful for the software company.

Total number of people = 15 + 75 = 90

The number of people who said yes = (2/5)90 = 36

The number of people who said no = 90 - 36 = 54

The number of men said yes = (6/15)x15 = 6

The number of men said No = 15 - 6 = 9

The number of women who said no = 54 - 9 = 45

Percentage = (45/90)x100

The percentage of women who said ‘no’ out of the total survey population:

= 50%

Thus, 50% of women who said ‘no’ out of the total survey population if 15 men and 75 women were asked if they had seen the company’s latest TV advertisement.

Learn more about the percentage here:

brainly.com/question/8011401

#SPJ1

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Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
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The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

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s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702

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*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}

     =176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

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