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Determine which is the appropriate approach to conduct a hypothesis test. ​Claim: The​ AI, or Adequate Intake of​ water, for pregnant women is a mean of​ 3L/d, liters per day. Sample​ data: nequals​200, x overbar equals​2.5, sequals1. The sample data appear to come from a normally distributed population with sigma equals1.2.

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SaniShahbaz

Answer:

One sample z-test for population mean would be the best approach to conduct a hypothesis test.

Step-by-step explanation:

Following is the data available to us:

Mean amount of water = u = 3

Sample mean = x = 2.5

Sample Size = n = 200

Sample Standard Deviation = s = 1

Population Standard Deviation = [tex]\sigma[/tex] = 1.2

Population is normally distributed.

We need to find the best approach to conduct a hypothesis test. Since only one sample is involved, it is a One-Sample test about population mean. for conducting hypothesis test for One-Sample about population mean we have following two options:

  1. One Sample z-test for population mean
  2. One sample t-test for population mean

Selecting the best approach:

The first thing to check is if our data from a population which is normally distributed. Which in this case is. Next we check if the value of population standard deviation is known or unknown. the rule is:

  • If value of population standard deviation is known, then we use One sample z-test
  • If value of population standard deviation is unknown and only value of sample standard deviation is known, then we use one-sample t-test.

Since, in this case we know the value of Population Standard Deviation which is 1.2, One sample z-test for population mean would be the best approach to conduct a hypothesis test.

The conclusion is that there is sufficient evidence to support the claim.

To understand the calculations, check below

Test Statistic:

In a test of hypothesis, the test statistic is a function of the sample data used to decide whether or not to reject the null hypothesis.

The formula for the test statistic is,

[tex]z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]

Given that,

Sample mean=2.4

Sample size=200

Population standard deviation=1.2

Level of significance=0.05

Critical values=[tex]\pm1.960[/tex]

Substituting the given values into the above formula we get,

[tex]z=\frac{2.5-3}{\frac{1.2}{\sqrt{200} } }\\ =-5.893[/tex]

And the two-tailed p-value by the z score table is 0.0000

Since the p-value is less than the level of significance.

So, reject [tex]H_0[/tex]

Learn more about the topic Test Statistic:

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