online photos: a poll surveyed 1765 internet users and found that 865 of them had posted a photo or video…

online photos: a poll surveyed 1765 internet users and found that 865 of them had posted a photo or video online. can you conclude that less than half of internet users have posted photos or videos online? use the $alpha = 0.01$ level of significance and the $p - value$ method with the ti - 84 plus calculator.\npart 1 of 5\n(a) state the appropriate null and alternate hypotheses.\n$h_0: p = 0.5$\n$h_1: p < 0.5$\nthis hypothesis test is a left - tailed test.\npart 2 of 5\n(b) compute the value of the test statistic. round the answer to two decimal places.\n$z=-0.84$\npart: 2 / 5\npart 3 of 5\n(c) compute the $p - value$. round the answer to at least four decimal places.\n$p - value=$

online photos: a poll surveyed 1765 internet users and found that 865 of them had posted a photo or video online. can you conclude that less than half of internet users have posted photos or videos online? use the $alpha = 0.01$ level of significance and the $p - value$ method with the ti - 84 plus calculator.\npart 1 of 5\n(a) state the appropriate null and alternate hypotheses.\n$h_0: p = 0.5$\n$h_1: p < 0.5$\nthis hypothesis test is a left - tailed test.\npart 2 of 5\n(b) compute the value of the test statistic. round the answer to two decimal places.\n$z=-0.84$\npart: 2 / 5\npart 3 of 5\n(c) compute the $p - value$. round the answer to at least four decimal places.\n$p - value=$

Answer

Explanation:

Step1: Recall the formula for P - value in left - tailed z - test

For a left - tailed z - test with test statistic $z$, the $P$-value is $P(Z < z)$. Here $z=- 0.84$.

Step2: Use the standard normal distribution table or calculator

We use the standard normal distribution (where $Z\sim N(0,1)$). Using a standard normal table or a calculator with a normal - distribution function (e.g., on a TI - 84 Plus: normalcdf(-1000,-0.84,0,1)), we find the probability. The $P$-value is the area to the left of $z =-0.84$ under the standard normal curve. $P(Z < - 0.84)=0.2005$

Answer:

$0.2005$