Question

The University of the Free State wishes to investigate if a new variety of wheat differs in resistant to diseases compared to an old variety. It is known that this disease strikes approximately 15% of all plants of the variety. A field experiment was conducted, and of 120 new plants, 20 become infected. Test at 1% level of significance if indeed the new variety of wheat differs in resistant to diseases compared to an old variety. State the null and alternative hypothesis

Answer 1

In this scenario, we want to test whether the new variety of wheat has a different rate of disease resistance compared to the established rate of 15% (0.15) for the old variety.

**Null Hypothesis (H0):** The proportion of infected plants in the new variety is equal to the proportion of infected plants in the old variety.
\[ H_0: p = 0.15 \]

**Alternative Hypothesis (H1):** The proportion of infected plants in the new variety is different from the proportion of infected plants in the old variety.
\[ H_1: p \neq 0.15 \]

Where \( p \) is the proportion of infected plants in the new variety of wheat.

We're conducting a two-tailed test since we're interested in differences in either direction (either higher or lower infection rates).

Next, we will perform a hypothesis test using the data provided.

1. **Sample Size (n)** = 120
2. **Number of infected plants (x)** = 20
3. **Sample Proportion (\(\hat{p}\))** = \(\frac{x}{n} = \frac{20}{120} = 0.1667\)
4. **Significance Level (\(\alpha\))** = 0.01

To calculate the test statistic, we will use the formula for the standard normal (Z) test:

\[
Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]

Where:
- \( p_0 \) is the hypothesized population proportion (0.15).
- \( \hat{p} \) is the sample proportion (0.1667).
- \( n \) is the sample size (120).

Plugging in the numbers:

\[
Z = \frac{0.1667 - 0.15}{\sqrt{\frac{0.15(1 - 0.15)}{120}}}
\]

Calculate the standard error:

\[
\sqrt{\frac{0.15 \times 0.85}{120}} = \sqrt{\frac{0.1275}{120}} \approx \sqrt{0.0010625} \approx 0.0326
\]

Now calculating \( Z \):

\[
Z = \frac{0.1667 - 0.15}{0.0326} \approx \frac{0.0167}{0.0326} \approx 0.511
\]

Next, we will compare the calculated Z-value with the critical Z-value for a two-tailed test at the 1% significance level. The critical Z-values corresponding to a significance level of 0.01 (0.005 in each tail) are approximately ±2.576.

Since \( |Z| = 0.511 < 2.576 \), we fail to reject the null hypothesis.

### Conclusion:
At the 1% level of significance, there is not enough evidence to conclude that the new variety of wheat differs in its resistance to diseases compared to the old variety.