Question

The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. As of August of $2012$, the standard deviation of Internet speeds across Internet Service Providers (ISPs) was $12.2$percent. Suppose a sample of $15$ISPs is taken, and the standard deviation is$13.2$. An analyst claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom, the test statistic, sketch the graph of the p-value, and draw a conclusion. Test at the$1\%$ significance level.

Short answer

The alpha value has been set at $0.01$. Now, because the $p-$ value is greater than $>\alpha$, the choice will be to reject the null hypothesis, $H_0$. As a result, the null hypothesis is not rejected. As a result, sufficient information exists to establish that the standard deviation of speeds is comparable to what was stated.

Step-by-step solution

Given information

Given in the question that, The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. As of August of$2012$, the standard deviation of Internet speeds across Internet Service Providers (ISPs) was $12.2$percent. Suppose a sample of $15$ISPs is taken, and the standard deviation is$13.2$. An analyst claims that the standard deviation of speeds is more than what was reported. we need to state the null and alternative hypotheses, compute the degrees of freedom, the test statistic, sketch the graph of the p-value, and draw a conclusion. And also test at the$1\%$ significance level.

Explanation

The null hypothesis is: $H_0$: The standard deviation of speeds is the same as it was previously reported.

y13y$13H_0:{\sigma }^2=(12.2{)}^2$, to be precise.

The alternative hypothesis is as follows:

$H_a$: The standard deviation of speeds is higher than reported.

Specifically,

$H_0:{\sigma }^2>(12.2{)}^2$

The following formula can be used to calculate degrees of freedom:

$df=n-1$

$\Rightarrow df=15-1$

$\Rightarrow df=14$

The following formula can be used to compute the test statistics:

Statistical test$=\frac{(n-1)s^2}{{\sigma }^2}$

The sample size is $n$, the sample variance is$s^2$, and the population variance is${\sigma }^2$. As a result, the computation is as follows:

$\text{Test statistic}=\frac{(n-1)s^2}{{\sigma }^2}$

$=\frac{(15-1)(13.2{)}^2}{(12.2{)}^2}$

$=\frac{14\times 174.24}{148.84}$

$=16.39$

In Excel, the p-value can be determined using the CHIDIST () formula, as illustrated below:

p-value$=0.29014$

p-value graph