A Food Marketing Institute found that 29% of households spend more than $125 a week on groceries. Assume the population proportion is 0.29 and a simple random sample of 207 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is less than 0.31?
There is a ___ probability that the sample proportion of households spending more than $125 a week is less than 0.31. Round the answer to 4 decimal places.

Respuesta :

Using the normal distribution, there is a 0.7357 = 73.57% probability that the sample proportion of households spending more than $125 a week is less than 0.31.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
  • By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex], as long as [tex]np \geq 10[/tex] and [tex]n(1 - p) \geq 10[/tex].

The estimate and the sample size are:

p = 0.29, n = 207.

Hence the mean and the standard error are given as follows:

  • [tex]\mu = p = 0.29[/tex].
  • [tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.29(0.71)}{207}} = 0.0315[/tex].

The probability that the sample proportion of households spending more than $125 a week is less than 0.31 is the p-value of Z when X = 0.31, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{0.31 - 0.29}{0.0315}[/tex]

Z = 0.63

Z = 0.63 has a p-value of 0.7357.

0.7357 = 73.57% probability that the sample proportion of households spending more than $125 a week is less than 0.31.

More can be learned about the normal distribution at https://brainly.com/question/4079902

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