Question

In January 2005, a company that monitors Internet traffic (WebSideStory) reported that its sampling revealed that the Mozilla Firefox browser launched in 2004 had grabbed a 4.6% share of the market. I. If the sample were based on 2,000 users, could Microsoft conclude that Mozilla has a less than 5% share of the market? II. WebSideStory claims that its sample includes all the daily Internet users. If that’s the case, then can Microsoft conclude that Mozilla has a less than 5% share of the market?

Answer 1

Answer:

(I) Mozilla has more than 5 percent or has equal to 5% share in the market.

(II) Yes, Mozilla has a less than 5% share of the market.

Step-by-step explanation:

It is given in the question that in the month of January 2005, WebSideStory, an organization which checks internet traffic had reported its sampling revealed which Mozilla Firefox browser had launched in the year 2004 has grabbed a share of 4.6% share of the global market.

(I) Now suppose the population proportion share of  market by the Mozilla  is = p

Then, Null Hypothesis is $H_0 \text{ is}\ p \geq 5$ %       {meaning Mozilla has more than 5 percent or equal to 5 percent share of the market}

Alternate Hypothesis, $H_A \text{ is}\ p < 5$ %       {meaning Mozilla has a less than five percent share of the market}

This test statistics which will be used is One-sample z-test for proportions;

$TS = \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}} \sim N(0,1)$

where,  $\hat{p}$ = is the sample proportion of share of the market that is grabbed by the Mozilla in year 2004 = 4.6%

n = sample of users = 2,000

So, the test statistics is  =   $= \frac{0.046-0.05}{\sqrt{\frac{0.05(1-0.05)}{2000}}}$

                                       = $-0.821$

Therefore,  z-test statistics is -0.821.

As per question it is not mentioned properly or stated the degree of significance, we can assume that it is 5%. Now, the level of significance at 5 percent, the z table will give the critical value of -1.96 to the left-tailed test.

Now the test statics value is greater than critical value of z, and thus we don't have sufficient evidence to reject the null hypothesis because it will not be placed in the rejection part or region.

So we can conclude that the Mozilla has equal to 5%  or more than 5 % share of the market.

(II) Now it is claimed by WebSideStory that their sample contains all the internet users using daily. Thus it means  4.6 percent share of market shows the entire population.

So, we conclude that the Mozilla has a share in the ,market of less than 5 percent.

Answer 2

Step-by-step explanation:

(I) Now suppose the population proportion share of market by the Mozilla is = p

Then, Null Hypothesis is H_0 \text{ is}\ p \geq 5H

0

is p≥5 % {meaning Mozilla has more than 5 percent or equal to 5 percent share of the market}

Alternate Hypothesis, H_A \text{ is}\ p < 5H

A

is p<5 % {meaning Mozilla has a less than five percent share of the market}

This test statistics which will be used is One-sample z-test for proportions;

TS = \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}} \sim N(0,1)TS=

n

p(1−p)

p

^

−p

∼N(0,1)

where, \hat{p}

p

^

= is the sample proportion of share of the market that is grabbed by the Mozilla in year 2004 = 4.6%

n = sample of users = 2,000

So, the test statistics is = = \frac{0.046-0.05}{\sqrt{\frac{0.05(1-0.05)}{2000}}}=

= -0.821−0.821

Therefore, z-test statistics is -0.821.

As per question it is not mentioned properly or stated the degree of significance, we can assume that it is 5%. Now, the level of significance at 5 percent, the z table will give the critical value of -1.96 to the left-tailed test.

Now the test statics value is greater than critical value of z, and thus we don't have sufficient evidence to reject the null hypothesis because it will not be placed in the rejection part or region.

So we can conclude that the Mozilla has equal to 5% or more than 5 % share of the market.

(II) Now it is claimed by WebSideStory that their sample contains all the internet users using daily. Thus it means 4.6 percent share of market shows the entire population.

So, we conclude that the Mozilla has a share in the ,market of less than 5 percent.