Careers in science: The General Social Survey asked 514 people whether they had ever considered a career in science, and 178 said that they had

Question

Careers in science: The General Social Survey asked 514 people whether they had ever considered a career in science, and 178 said that they had. Can you conclude that more than 30% of people have considered a career in science? Use the α = 0.05 level of significance.

Answer 1

In this case you are using the central limit theorem to approximate a normal distribution for a proportion.

So p = y/n, where y is the number of people who said yes and n is the number of people who said yes.

So p = 178/514 = .346

We want to check if p is > .3., where we say p0 = .3.

So our null hypothesis is that p <= .3 and our alternative hypothesis is p >.3

so to get our test statistics we do (p - p0)/sqrt((p0*(1-p0))/n) = (.346-.3)/sqrt((.3*(1-.3))/514)

so Z = .046/0.0202 = 2.28, so now we can calculate our p-value. the definition of our p-value is the probability of observing values like the values from the data or more extreme. So that would mean since our alternative hypothesis is p > .3, then we want P(Z>2.28) = 0.0113 = p-value.

Now we compare our p-value to our level of significance which is 0.05. Since our p-value = 0.0113 which is less than our alpha of 0.05 we reject the null hypothesis. This suggest that there is evidence to support the proportion of people who consider a career in science is greater than 30%.

Answer 2

ho : 30% or less of people have considered a career in science, or p₁ ≤ 0.3

h1 : more than 30% of people have considered a career in science, or p₁ > 0.3

α = 0.05

We must conduct a 1 proportion z test using the following formula:

Ζ = (p_{1 -} p₀) / √((p₀(1-p₀))/n)

*I understand this notation is difficult to read, so i recommend double checking your formula here:

https://www.tutorialspoint.com/statistics/one_proportion_z_test.htm

First, we must find the observed proportion value:

p₁ = x/n = 178/514 = 0.346

So, our values are:

p₀ (null hypothesized value) = 0.3

p₁ (observed value) = 0.346

n (sample size) = 514

Next, we plug in our values to the equation, and solve for the Z test statistic. If solved correctly, you should get a test statistic value of Z = 2.29

Finally, either by using a statistic calculator or a z table, find the p value of Z = 2.29, which is 0.011

Conclude that because our p value of 0.011 is less than the alpha value of 0.05, we reject the null hypothesis and conclude that more than 30% of people have considered a career in science

Answer 3

The method I'm going to use here relies on the classical frequentist model of statistics. There are other approaches, which, believe it or not, give other answers, but the classical frequentist model is the most widely used.

The method is as follows:

1. Assume for the moment that exactly 30% of the population have considered careers in science.
2. Ask "if the assumption in part 1 were true, what is the probability that at least 178 out of 514 people would say they were interested in a career in science?" Calculate that probability and call it p.
3. If p is less than our α level, (in this case α=0.05), then we conclude that more than 30% of the population is interested in a career in science.

The only hard part of the calculation is part 2). If exactly 30% of the population were interested in a career in science, what would happen if we asked 514 people? We don't know exactly, because there's some randomness depending on which people we happened to ask. However, we can calculate the mean and standard deviation of the number of "yes" answers.

The mean can be calculated

μ=0.30*514 = 154.2

We don't care that the mean is not a whole number, this is perfectly fine. The standard deviation can be calculated

σ = sqrt(0.3*0.7*514) = 10.39

Next, we ask "what is the probability that we'd get at least 178 "yes" answers?" We want to use the normal distribution. The one hitch is that the normal distribution can take any number, but the number of people who answer "yes" must be a whole number, so we use rounding for the normal distribution. Thus, to get at least 178 "yes" answers, we need the normal distribution to give us an answer of at least 177.5.

For a normal distribution with mean 154.2 and standard deviation 10.39 to give us a value of 177.5, we need a z-score of

z = (177.5 - 154.2)/10.39 = 2.24

Using a normal distribution table, we find that a z-score this high will only happen around 1% of the time. Because 1% is less than our α level of 5%, we conclude that it is likely that the true proportion of people interested in a career in science is more than 30%.

Answer 4

You can perform a 1-PropZTest to answer this question. To find this on a TI-84 (or similar) calculator, press STAT then use the right arrow to go over to TESTS.

For this problem,

the null hypothesis is that p=.30

the alternative hypothesis is that p>.30

x=178 people (who said they had considered a science career)

n=514 people (who were asked)

When you plug in all of the values, you will get a p value. In this scenario, you get p=0.011. The question asked to use α = 0.05. Since p<α, you would reject the null hypothesis. Therefore, there is sufficient evidence to support the claim that more than 30% of people have considered a career in science.