hypothesis testing

Question

Microsoft estimated tht out of 10000 potential software buyers, 35% are planning to purchase the new software until and upgrade has been released. After an advertising campaign.Microsoft survey 3000 buyers and found 950 who are still skeptical. At 5% level significance.can the company concluded tht the population of skeptical population has decreased.

Answer 1

35% of 10,000=3,500 (mean₁). That's Sample 1 where n₁=10000. 65% (6,500) are still skeptical.

950 (mean₂) out of 3000 is 950/3000=19/60 (31.67%) are skeptical (do not accept the new software). That's Sample 2 where n₂=3000. In this sample, 41/60 (68.33%) accept the new software.

Both of these are large samples so the Normal Distribution can be assumed.

This is a binary situation and variance=np(1-p) where p is the proportion (%ge).

variance₁=σ₁²=10000×0.35×0.65=2275; SD₁=σ₁=47.70;

variance₂=σ₂²=3000×41/60×19/60=649.17; SD₂=σ₂=25.48.

We need to compare like with like, so let's consider the difference in mean proportions who are willing to accept the new software until an upgrade becomes available.

0.6833-0.35=0.3333=⅓.

Now we need a SD σ to generate a test statistic. σ=√(σ₁²/n₁+σ₂²/n₂)=√(2275/10000+649.17/3000)=1.58 approx.

The test statistic Z=0.33/1.58=0.21 approx. This is the number of SDs deviation from the mean difference of the means of the two samples.

At 5% significance Z=1.96 so 0.21 is considerably less than 1.96 so lies within the range of what is probably (with 95% degree of confidence) not significant.

The company can conclude that the skeptical population has not decreased significantly.