Ahmed is going to catch the bus that leaves at 8.16 am. This bus takes 25 minutes to

In an examination of purchasing patterns of shoppers, a sample of 16 shoppers revealed that they spent, on average, $54 per hour

Lesechka [4]

Answer:

$54-1.64\frac{21}{\sqrt{16}}=45.39$

$54 +1.64\frac{21}{\sqrt{16}}=62.61$

So on this case the 90% confidence interval would be given by (45.39;62.61)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=54$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=21$ represent the sample standard deviation

n=16 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.90 or 90%, the value of $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$

Now we have everything in order to replace into formula (1):

$54-1.64\frac{21}{\sqrt{16}}=45.39$

$54 +1.64\frac{21}{\sqrt{16}}=62.61$

So on this case the 90% confidence interval would be given by (45.39;62.61)

4 0

3 years ago

The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 3639 3639 miles, with a var

Andrei [34K]

Answer:

96.6% probability that the mean of a sample would differ from the population mean by less than 126 miles

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

A reminder is that the standard deviation is the square root of the variance.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .