A medium artichoke contains about 14% of the recommended amount of a certain mineral an average adult should have each day. Abo
1 answer:
Answer:
0.0238 g
Step-by-step explanation:
Given that:
the certain mineral = potassium and a medium artichoke contain about 0.17 g of it.
Suppose a medium artichoke contains about 14% of the recommended amount of a certain mineral.
On Average, the amount an adult should have each day = 0.17 g × (14/100)
= 0.0238 g
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Answer:
We conclude that the majority of people prefer the taste of fresh-brewed coffee.
Step-by-step explanation:
We are given that of the 50 subjects who participate in the study, 16 prefer the instant coffee.
We have to test the claim that a majority of people prefer the taste of fresh-brewed coffee.
<em />
<u><em>Let p = proportion of the population who prefer fresh-brewed coffee</em></u>
So, Null Hypothesis, : p
50% {means that the majority of people does not prefer the taste of fresh-brewed coffee}
Alternate Hypothesis, : p > 50% {means that the majority of people prefer the taste of fresh-brewed coffee}
The test statistics that would be used here <u>One-sample z proportion statistics;</u>
T.S. = ~ N(0,1)
where, = sample proportion who prefer fresh-brewed coffee =
= 0.68
n = sample of subjects = 50
So, <u><em>test statistics</em></u> =
= 2.73
The value of z test statistics is 2.73.
<u>Now, P-value of the test statistics is given by;</u>
P-value = P(Z > 2.73) = 1 - P(Z 2.73)
= 1 - 0.99683 = 0.0032
Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of 1.645 for right-tailed test.
<em>Since our test statistics is more than the critical value of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>
Therefore, we conclude that the majority of people prefer the taste of fresh-brewed coffee.