According to a marketing research study, american teenagers watched 16.5 hours of social media posts per month last year, on average. a random sample of 20 american teenagers was surveyed and the mean amount of time per month each teenager watched social media posts was 17.3. this data has a sample standard deviation of 2.1. (assume that the scores are normally distributed.)

researchers conduct a one-mean hypothesis at the 10% significance level to test if the mean amount of time american teenagers watch social media posts per month is greater than the mean amount of time last year.

which answer choice shows the correct null and alternative hypotheses for this test?

select the correct answer below:

h0: μ=17.3; ha: μ< 17.3, which is a left-tailed test.

h0: μ=17.3; ha: μ> 17.3, which is a right-tailed test.

h0: μ=16.5; ha: μ< 16.5, which is a left-tailed test.

h0: μ=16.5; ha: μ> 16.5, which is a right-tailed test.

H0:μ=16.5; Ha:μ>16.5, which is a right-tailed test.

Step-by-step explanation:

Let be the population mean .

Given : According to a marketing research study, American teenagers watched 16.5 hours of social media posts per month last year, on average.

hours

A random sample of 20 American teenagers was surveyed and the mean amount of time per month each teenager watched social media posts was 17.3.

i.e. 17.3 is the sample mean for 20 randomly sampled  American teenagers .

This data has a sample standard deviation of 2.1.

Researchers  want to test mean amount of time American teenagers watch social media posts per month is greater than the mean amount of time last year.

i.e. hours

So the appropriate hypothesis for this situation will be :

  [Null hypothesis takes '= 'sign ]

  [Alternative hypothesis takes >, < , ≠  signs]

Since alternative hypothesis right-tailed , so the hypothesis test is a right tailed test.

x = 17

step-by-step explanation:

1st step: isolate x.

24 - 3x = -27

-24         -24

-3x   =   -51

2nd step: divide.

-3x     =   -51

  3             3

  x       =     17

hope this