Question

A random sample of 64 bags of white cheddar popcorn weighed, on average, 5.23 ounces with a standard deviation of 0.24 ounce.
Test the hypothesis that µ = 5.5 ounces against the alternative hypothesis, µ < 5.5 ounces, at the 0.05 level of significance.

Answer 1

Answer:

We conclude that cheddar popcorn weighed less than 5.5 ounces.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ =5.5 ounces

Sample mean, $\bar{x}$ = 5.23 ounces

Sample size, n = 64

Alpha, α = 0.05

Sample standard deviation, σ = 0.24 ounce.

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 5.5\text{ ounces}\\H_A: \mu < 5.5\text{ ounces}$

We use One-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{5.23 - 5.5}{\frac{0.24}{\sqrt{64}} } = -9$

Now, $z_{critical} \text{ at 0.05 level of significance } =-1.645$

Since,  

$z_{stat} < z_{critical}$

We reject the null hypothesis and accept the alternate hypothesis. Thus, we conclude that cheddar popcorn weighed less than 5.5 ounces.

Answer 2

The z test is mathematically given as z=-1.86, and a>p value.

therefore we failed to accept null hypothesis.

What conclusion do we come to at the test of the hypothesis?

Question Parameters:

A random sample of 64 bags

weighed, on average, 5.23 ounces

the standard deviation of 0.24 ounce.

µ = 5.5 ounces

Generally, hypothesis  is mathematically given as

H_0:\mu = 5.8 null

H_a:\mu < 5.8 alter

Using z test

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}\\\\z=\frac{5.74-5.8}{\frac{0.26}{\sqrt{65}}}$

z=-1.86

In conclusion,

p value = 0.0314

α = 0.10

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