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Answer:
a) The percentage of households in the town with three or more largescreen TVs is estimated as :
The best estimation for the population proportion is :
And that represent the 1.4%.
b) And the 95% confidence interval would be given (0.00370;0.0243).
And the % would be between 0.37% and 2.43%.
Step-by-step explanation:
Data given and notation
n=500 represent the random sample taken
X=7 represent the households with three or more large-screen TVs
estimated proportion of households with three or more large-screen TVs
represent the significance level (no given, but is assumed)
z would represent the statistic (variable of interest)
p= population proportion of households with three or more large-screen TVs
Part a
The percentage of households in the town with three or more largescreen TVs is estimated as :
The best estimation for the population proportion is :
And that represent the 1.4%.
Part b
Yes is possible. We hav that and
so we have the assumption of normality to find the interval.
The confidence interval would be given by this formula
For the 95% confidence interval the value of and
, with that value we can find the quantile required for the interval in the normal standard distribution.
And replacing into the confidence interval formula we got:
And the 95% confidence interval would be given (0.00370;0.0243).
And the % would be between 0.37% and 2.43%.
Answer:
11.51% probability that the mean annual salary of the sample is between $71000 and $73500
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
What is the probability that the mean annual salary of the sample is between $71000 and $73500?
This is the pvalue of Z when X = 73500 subtracted by the pvalue of Z when X = 71000. So
X = 73500
By the Central Limit Theorem
has a pvalue of 0.1151
X = 71000
has a pvalue of 0.
0.1151 - 0 = 0.1151
11.51% probability that the mean annual salary of the sample is between $71000 and $73500