A study in 2010 reported that 3 in 4 adults in a relationship met their significant other using an online dating site. Recently a popular magazine 'People' claimed that that number has changed. To test this claim a telephone survey of 600 randomly selected adults in a relationship was completed. Of the 600 adults, 480 said that they met their significant other using an online dating site. Does this provide evidence that the proportion has changed? Use α = 0.05, to test the magazine's claim. Conditions: 1. The sample is 2. n p 0 ( 1 − p 0 ) 10 3. n 0.05 N Hypotheses: H 0 : p 0.75 H 1 : p 0.75 Test Statistic: The test statistic is a test statistic. The value of the test statistic is . p-value: The value of the p-value is . Decision: Because the p-value is than α = 0.05, we the null hypothesis. Conclusion: The data the claim that the proportion has changed

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

p = 3/4 = 0.75

For the alternative hypothesis,

p ≠ 0.75

Considering the population proportion, probability of success, p = 0.75

q = probability of failure = 1 - p

q = 1 - 0.75 = 0.25

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 480

n = number of samples = 600

P = 480/600 = 0.8

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

z = (0.8 - 0.75)/√(0.75 × 0.25)/480 = 2.53

Recall, population proportion, P = 0.75

The difference between sample proportion and population proportion(P - p) is 0.8 - 0.75 = 0.05

Since the curve is symmetrical and it is a two tailed test, the p for the left tail is 0.75 - 0.05 = 0.7

the p for the right tail is 0.75 + 0.05 = 0.8

These proportions are lower and higher than the null proportion. Thus, they are evidence in favour of the alternative hypothesis. We will look at the area in both tails. Since it is showing in one tail only, we would double the area

From the normal distribution table, the area above the test z score in the right tail 1 - 0.9943 = 0.0057

We would double this area to include the area in the left tail of z = - 2.53. Thus

p = 0.0057 × 2 = 0.0114

Because alpha, 0.05 > than the p value, 0.0114, then we would reject the null hypothesis.

Therefore, at 5% significance level, this data provide evidence that the proportion has changed.

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3