Question

In a random sample of 1000 American households, 37 watched the television program Young Sheldon on one particular Thursday. Is there enough evidence at a = 0.05 to support the claim that more than 3% of American households watched this program? Standardized test statistic: (Round to 2 or 3 decimal places.) P-value: (Round to 4 decimal places.) There (type is or is not) enough evidence at the 5% level of significance to support the claim that more than 3% of American households watched Young Sheldon.

Answer 1

Here we have given that,

n=number of American households=1000

x: number of American households watched the television program young Sheldon on one particular Thursday = 37

Now, we estimate the sample proportion $\hat p$ as

$\hat p$ =sample proportion American households watched the television program young Sheldon on one particular Thursday

=$\frac {x}{n}$

= $\frac {37}{1000}$

= 0.037

p=populaiton proportion of American households watched young sheldon= 3% =0.03

The below mentioned necessary assumption is satisfied for this hypothesis test (one sample proportion test).

• The data are simple random sample values form the population.
• The response is binary i.e. either success/failure and the distribution is binomial.
• The mean n*p=1000*0.03=30 and the variance = n*p*(1-p)= 1000*0.03*(1-0.03)=29 are greater than 10, the binomial distribution can be approximated by the normal distribution.

Claim: To check whether the proportion of American households watched young Sheldon more than 3% i.e. 0.03.

The null and alternative hypothesis are as follows,

v/s

where p is the population proportion of American households watched young Sheldon

This is the right one-tailed test.

Now, we can find the test statistic is as follows,

Z-statistics= $\frac{\hat p - p} {\sqrt \frac{p*(1-p)}{n}}$

   = $\frac{0.037 - 0.03} {\sqrt \frac{0.03*(1-0.03)}{1000}}$

=1.30

The test statistics is 1.30.

Now we find the P-value,

p-value=P(Z > z-statistics)

             =1- P( Z < 1.30)

             =1 - 0.90320 Using standard normal z table see the value corresponding to the z=1.30

              = 0.0968

The p-value is 0.0968

Decision:

$\alpha$ = level of significance= 0.05

Here p-value (0.0968) greater than (>) 0.05 $\alpha$

Conclusion:

We fail to reject the Ho (Null Hypothesis)

There is not sufficient evidence to support the claim that more than 3% of American households watched young sheldon.