Question

6. A new app start-up claims that "no more than than 6% of the product reviews they receive are negative" but we suspect it is more. When examining a random sample of 2000 reviews, we find that 140 are negative Let p be the probability that a review is negative and let X, (i 1,.. ,2000) be the random variable that represents the ith review being negative (assume they are independent and identically distributed). (a) Identify an appropriate probability model for X, and give its mean and 2 marks variance ơ2 [1 mark (b) What is a sensible estimator for p? c) Write down the null and alternative hypothesis to test whether the probability of a review being negative is greater than 0.06 or not. 2 marks 6 (Continued) d) Write down the test statistic and its distribution under the n hypoth 2 marks esis. (e) Write down the observed test statistic and calculate the P-value based on the 2 marks] data (f) At a a 0.05 significance level, do you reject the hypothesis test or not? 1 mark (g) Describe in words your decision regarding the claim that only 6% of reviews 2 marks are negative"

Answer 1

This is a case of Binomial distribution since Xi represents i^th review being negative and will assume 2 values :0 and 1 , 1 representing the review being negative

Now Xi~N(n,p)

where n=2000 and p=140/2000=0.07

Now since N is large we will assume it to follow normal distribution with mean np and variance np(1-p)

Therefore :

a)Xi~N(140,130.2)

b) The appropriate estimator p will be :

-2000 X.

c)

The null and alternate hypothesis is given as:

Ho 0.06 Hi D0.06

Further parts are done after hypothesis testing:

The following information is provided: The sample size is N = 2000, the number of favorable cases is X 140, and the sample proportion is 00.07, and the significance level is a 0.05 The following null and altemative hypotheses need to be tested: Ho:p=0.06 Ha:p>0.06 This corresponds to a right-tailed test, for which a z-test for one population proportion needs to be (2) Rejection Region Based on the information provided, the significance level is a 0.05, and the critical value for a right-tailed test is 1.64. The rejection region for this right-tailed test is R1.64) (3 The z-statistic is computed as follows 0,07-0.061.883 p Po ˇ pr (1-p)/n = V 0.06( 1-0.06)/2000

(4) Decision about the null hypothesis Since it is observed that 1.8831.64, it is then concluded that the null hypothesis is rejected Using the P-value approach: The p-value is p = 0.0298 , and since p concluded that the null hypothesis is rejected. 0.0298 < 0.05, it is (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is greater than po, at the a 0.05 significance level.

d)

The test statistics is given as:

P-Po Po (1

e)

The test statistics is:1.883 (calculated above)

p-value:0.0298

f)At alpha =0.05 we reject the null hypothesis because the p-value is less than 0.05.

g) Therefore we can conclude that the more than 6% of claim reviews are negative.(Since the null hypothesis is rejected)