Question

iPad 11:04 AM 59% (1 point) An internet service provider advertises that customers on its "high speed" plan will experience average download speeds of 20 Mbps. A dissatisfied customer believes his download speeds are significantly less than advertised. To test this claim, he checks his download speed at 49 randomly selected times over the course of one month. Assume this can be treated as a Simple Random Sample. Grades 1. Let u be the true average download speed (in Mbps) for this customer. An appropriate null hypothesis Problem Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 is: OAp < 20 OB, ? = 20 OD.? > 20 2. Let u be the true average download speed (in Mbps) for this customer. An appropriate alternative hypothesis is: Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 °C, ? > 20 3. In the sample data, download speeds had a mean of 19.4 Mbps and a standard deviation of 3.25 ii(round to at least three Mbps. The value of the test statistic for this sample is decimal places). 4. The P-value for this sample is (round to at least three decimal places, and do not convert to percent). 5, According to the results above and using a 5% significance level: A. The sample provides sufficient evidence to conclude that this customer's average download OB. The sample provides sufficient evidence to conclude that this customer's average download C. The sample does not provide sufficient evidence to conclude that this customer's average 0 D. The sample provides sufficient evidence to conclude that this customer's average download speed is less than advertised. speed is higher than advertisec download speed is less than advertised. speed is exactly as advertised.

Answer 1

Given : n = 49    $\bar{X}=19.4$    s = 3.25    $\mu_0=20$    $\alpha=0.05$

Hypothesis :

1 . Null Hypothesis : $H_0:\mu=20$

2 . Alternative Hypothesis : $H_a:\mu<20$

3 . Since , the population standard deviation is unknown. Therefore , use t-test.

The test statistic under Ho is ,

$t=\frac{\bar{X}-\mu_0}{\sigma/\sqrt{n}}\to t_{n-1}$

$=\frac{19.4-20}{3.25/\sqrt{49}}=-1.292$

4 . P value = $P(t_{d.f.}>|t_{stat}|)=P(t_{49-1}>1.292)=0.013$

5 . Decision : Here , P value = 0.013 < $\alpha=0.05$

Therefore , reject Ho at 5% level of significance

Hence , The sample provides sufficient evidence to conclude that this customer's average download speed is less than advertised.