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A proportion test is a statistical tool used when you want to compare a sample proportion to a known proportion or to compare two sample proportions. In this exercise, we examine whether the proportion of television sets needing repair exceeds the previous experience rate of 10%.

The proportion test aims to determine if the observed sample proportion, in this case, 18% of the TV sets needing repair, indicates a significant change from the expected 10%. This involves setting up a null hypothesis ( ull hypothesis ) and an alternative hypothesis ( alternate hypothesis ):

• Null Hypothesis ( ull hypothesis ): The proportion of sets needing repair is 10% or less.
• Alternate Hypothesis ( alternate hypothesis ): The proportion of sets needing repair is greater than 10%.

Based on the outcome of the test statistic and the critical value, we decide whether there is enough evidence to reject the null hypothesis.

The significance level, denoted by \( \alpha \), is a threshold used to decide whether to reject the null hypothesis. In this exercise, the significance level is set at 0.05. This signifies a 5% risk of concluding that the proportion of sets needing repair has increased when it hasn't.

Choosing a significance level involves a tradeoff:

• A lower \( \alpha \) decreases the chances of a Type I error (incorrectly rejecting the null hypothesis), but may increase the chance of a Type II error (failing to detect an actual effect).
• A higher \( \alpha \) increases the risk of a Type I error but makes it easier to detect an actual effect.

The significance level is especially crucial in hypothesis testing, guiding the researcher in making informed decisions about rejecting or not rejecting the null hypothesis.

The \( p \)-value in hypothesis testing helps determine the strength of the results. It's a measure of the probability of observing a test statistic as extreme as, or more extreme than, the observed value, assuming the null hypothesis is true.

In our case, the calculated \( z \)-value was 1.89. Using a standard normal distribution (Z-table), you find the area to the right of this \( z \)-value to get the \( p \)-value.

• The smaller the \( p \)-value, the stronger the evidence against the null hypothesis.
• If the \( p \)-value is less than the significance level (\( \alpha = 0.05 \)), we reject the null hypothesis.
• In this exercise, we found a \( p \)-value of approximately 0.029, which is less than our significance level of 0.05, supporting the rejection of the null hypothesis.

The \( p \)-value provides a quantitative measure for decision making, highlighting whether the evidence is strong enough to make a conclusion about the population proportion.

A Z-test is a type of hypothesis test used for situations where the data can be described using statistics such as a mean. For proportions, like in the exercise, we use the Z-test to compute whether the observed sample proportion significantly differs from a specified value.

In this scenario, the Z-test formula for a sample proportion is: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1-p_0)}{n}}} \]

• \( \hat{p} \) is the sample proportion (0.18).
• \( p_0 \) is the hypothesis proportion (0.10).
• \( n \) is the sample size (50).

Plugging in these values gives a \( z \)-value of 1.89. This value is compared to the critical \( z \)-value to determine statistical significance:

• If the calculated \( z \)-value is greater than the critical \( z \)-value (for \( \alpha = 0.05 \), critical \( z \)-value is about 1.645 for a right-tailed test), the null hypothesis is rejected.

Using the Z-test, we concluded that there is substantial evidence that the proportion of television sets needing repair has increased beyond the original expectation.