Question

1. You and a group of friends wish to start a company. You have an idea, and you are comparing startup incubators to apply to. (Startup incubators hold classes and help startups to contact venture capitalists and network with one another) Assume funding is normally distributed.

Incubator A has an 80% success ratio getting companies to survive at least 4 years from inception. The average venture funding of the 28 companies reaching that 4-year mark, is 1.3 million dollars with a standard deviation of 0.6 million

Incubator B has a 60% success ratio getting companies to survive at least 4 years from inception. The average venture funding of the 21 companies reaching that 4-year mark, is 1.9 million dollars with a standard deviation of 0.55 million

a. Are the success ratios significantly different? (note the count is given but not N, how do you find N?)

b. Is the average funding in incubator B significantly different from the average funding in a? (use a=0.01). Assume a normal distribution

Answer 1

a.

For company A, the sample proportion is

For company B, the sample proportion is

The value of the pooled proportion is computed as

Also, the given significance level is α=0.05.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:p1​=p2​.

Ha:p1​̸​=p2​

This corresponds to a two-tailed test, for which a z-test for two population proportions needs to be conducted.

(2) Rejection Region

The significance level is α = 0.05, and the critical value for a two-tailed test is .

The rejection region for this two-tailed test is R = { z : ∣z∣ > 1.96}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that ∣z∣ = 3.086 > zc ​= 1.96, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0.002, and since p=0.002<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population proportion p1​ is different than p2​, at the 0.05 significance level.

Therefore, there is enough evidence to claim that the success ratios significantly differ of companies A and companies B.

b.

1 represent incubator A and 2 represent incubator B

The sample means are shown below:

Also, the sample standard deviations are:

and the sample sizes are n1​=28 and n2​=21.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ ≠ μ2​

μ1 represent average funding in incubator A and μ2 represent average funding in incubator B

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

The critical values are FL​=0.444 and FU​=2.375, and since F = 1.19, then the null hypothesis of equal variances is not rejected.

(2) Rejection Region

The significance level is α=0.05, and the degrees of freedom are df = 47. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is tc ​= 2.684556, for α = 0.01 and df = 47.

The rejection region for this two-tailed test is R = { t : ∣t∣ > 2.684556}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that ∣t∣ = 6.195 > tc ​= 2.684556, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p =1.363351e-07 0, and since p = 0 < 0.01, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is different than μ2​, at the 0.01 significance level.

Therefore, there is enough evidence to claim that average funding in incubator B significantly different from the average funding in A at the 0.01 significance level.

Confidence Interval

The 95% confidence interval is−0.795<μ1​−μ2​<−0.405.