Question

The display provided from technology available below results from using data for a smartphone​ carrier's data speeds at airports to test the claim that they are from a population having a mean less than 6.00 Mbps. Conduct the hypothesis test using these results. Use a 0.050 significance level. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim.

null? hypotheses?

z= ??

p-vale?

conclusion? evidence or not enough evidence?

Answer 1

As the sample data is not given, i will assume that the sample mean is x and sample standard deviation is s and sample size is n. Now claim is that they are from a population having mean less than 6.00 Mbps.

Thus null hypothesis H0: x<=6 (Mean is less than 6 mbps), Alternate hypothesis is H1: x>6.

Here SE = s/sqrt(n).

Z-score = x-6/SE.

p-value = 1-pnorm(x,6,s) #Use R for using this function.

Now at 95% confidence or 0.05 significance Z-critical is 1.96, since it is a one-sided test. We will reject H0 if z>1.96.

So if Z-score is greater than 1.96 we have enough evidence to tell that sample comes from population which has mean greater than 6.

Kindly attach the data for exact calculation.