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A market research firm conducts telephone surveys with a 44% historical response rate

A market research firm conducts telephone surveys with a 44% historical response rate. What is the probability that in a new sample of 400 telephone numbers, at least 150 individuals will cooperate and respond to the questions? In other words, what is the probability that the sample proportion will be at least 150/400 = .375?
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Answer #1

General guidance

Concepts and reason

Sampling Distribution of sample Proportion:

The random sample of size n is taken from the population with sample proportionp^\\hat p.

The sampling distribution of the sample proportion has mean pp and the standard deviationp(1\u2212p)n\\sqrt {\\frac{{p\\left( {1 - p} \\right)}}{n}} . Moreover, the sample proportion follows normal distribution for large sample size n.

Standard Error:

The standard error of the mean is defined as the value of the standard deviation of all possible sample means for the given sample size.

Margin of Error:

The margin of error is defined as a statistic which gives the amount of sampling error in the given study. Also, the margin of error tells the percentage of points that the obtained results would differ from that of the given population value. The half length of the confidence interval is also known as margin of error.

Sample size:

In statistics, the sample size is defined as the number of subjects included in a sample or it is a group of observations which is coming from the population and is considered a representative of the true population.

Confidence interval:

A range of values such that the population parameter can expected to contain for the given confidence level is termed as the confidence interval. In other words, it can be defined as an interval estimate of the population parameter which is calculated for the given data based on a point estimate and for the given confidence level.

Moreover, the confidence level indicates the possibility that the confidence interval can contain the population parameter. Usually, the confidence level is denoted by . The value is chosen by the researcher. Some of the most common confidence levels are 90%, 95%, and 99%.

Fundamentals

Formula for Z-score of sample proportion is given below:

Z=p^\u2212pp(1\u2212p)nZ = \\frac{{\\hat p - p}}{{\\sqrt {\\frac{{p\\left( {1 - p} \\right)}}{n}} }}

Where p is the population proportion and p^\\hat pis the sample proportion.

The standard deviation of the sampling distribution of the sample proportion is given below:

\u03c3p^=p(1\u2212p)n{\\sigma _{\\hat p}} = \\sqrt {\\frac{{p\\left( {1 - p} \\right)}}{n}}

Central limit theorem for the sample proportion:

\uf0b7In the case of the population proportion p, the sampling distribution of p^\\hat pcan be approximated to normal, if the sample size is large.

The formula for standard error is,

SE=\u03c3nSE = \\frac{\\sigma }{{\\sqrt n }}

The formula of margin of error is,

E=Z\u03b12.\u03c3nE = \\frac{{{Z_{\\frac{\\alpha }{2}}}.\\sigma }}{{\\sqrt n }}

The formula of sample size is,

The formula of margin of error when confidence interval is given,

First Step | All Steps | Answer Only

Step-by-step

Step 1 of 5

(1)

From the given information, the sample proportion is 0.375, population proportion (p) is 0.40 and the sample size (n) is 400.

The probability that the sample proportion will be at least 0.375 is given below:

From the \u201cstandard area normal table\u201d the area to the left of is 0.1539.

Thus,

Part 1

The probability that the sample proportion will be at least 0.375 is 0.8461.


Explanation | Hint for next step

The probability that the sample proportion will be at least 0.375 is obtained by subtracting the area to the left of from the total probability 1. It is expected that 84.61% of times the sample proportion would be at least 0.375.

Step 2 of 5

(2.a)

From the given information the standard deviation(\u03c3)\\left( \\sigma \\right) of normal distribution is 4. The sample size (n) is 77.

The standard error of mean is obtained below:

SE=\u03c3n=477=48.77=0.46\\begin{array}{c}\\\\SE = \\frac{\\sigma }{{\\sqrt n }}\\\\\\\\ = \\frac{4}{{\\sqrt {77} }}\\\\\\\\ = \\frac{4}{{8.77}}\\\\\\\\ = 0.46\\\\\\end{array}

Part 2.a

The standard error of the mean is 0.46.


Explanation | Hint for next step

The standard error of mean is obtained by dividing the standard deviation by the square root of the sample size.

Step 3 of 5

(2.b)

The Z\u03b12{Z_{\\frac{\\alpha }{2}}}value is obtained by using standard normal table as shown below:

Consider, the confidence level is 0.95.

For(1\u2212\u03b1)=0.95\u03b1=0.05\u03b12=0.025\\begin{array}{c}\\\\{\\rm{For}}\\left( {1 - \\alpha } \\right) = 0.95\\\\\\\\\\alpha = 0.05\\\\\\\\\\frac{\\alpha }{2} = 0.025\\\\\\end{array}

From standard normal table, the required Z0.025{Z_{0.025}} value for 95% confidence level is 1.96.

The margin error with 95% confidence is obtained below:

E=Z\u03b12\u03c3n=1.96\u00d7477=7.848.77=0.89\\begin{array}{c}\\\\E = \\frac{{{Z_{\\frac{\\alpha }{2}}}\\sigma }}{{\\sqrt n }}\\\\\\\\ = \\frac{{1.96 \\times 4}}{{\\sqrt {77} }}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\ = \\frac{{7.84}}{{8.77}}\\\\\\\\ = 0.89\\\\\\end{array}

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