A manufacturer of sports equipment claims that the new fishing line has a mean breaking strength of at least 8 kilograms. Test this claim if a sample of 28 lines found that the mean breaking strength is 7.8 Kg with sample standard deviation 0.5 Kg. The distribution is approximately normal.

State the null hypothesis Ho and the alternative hypothesis Ha. Give the test statistic, the p-value and the conclusion. Justify your answers completely. Is the result significant at the 5% level?

The null hypothesis (Ho) in this case would be that the mean breaking strength of the new fishing line is equal to 8 kilograms. The alternative hypothesis (Ha) would be that the mean breaking strength is less than 8 kilograms.

Ho: μ = 8
Ha: μ < 8

To test this claim, we can use a one-sample t-test since we have a sample mean, sample size, and sample standard deviation. The test statistic is calculated as:

t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))

Plugging in the values given:
t = (7.8 - 8) / (0.5 / sqrt(28))
= -0.2 / (0.5 / 5.291)

Calculating:
t ≈ -0.2 / 0.0946
≈ -2.114

To determine the p-value, we compare the calculated t-value with the t-distribution table for 27 degrees of freedom (sample size minus 1) at a one-tailed significance level of 0.05.

By referring to the t-distribution table, we find that the p-value for t = -2.114 is approximately 0.021.

Since the p-value is less than the significance level of 0.05, we can reject the null hypothesis. This means that there is enough evidence to suggest that the mean breaking strength of the new fishing line is less than 8 kilograms.

In conclusion, the result is significant at the 5% level, and we have enough evidence to support the manufacturer's claim that the new fishing line may not have a mean breaking strength of at least 8 kilograms.

To test the manufacturer's claim, we can use a one-sample t-test to compare the sample mean breaking strength with the claimed mean breaking strength.

Null hypothesis (H0): The mean breaking strength of the new fishing line is at least 8 kilograms.
Alternative hypothesis (Ha): The mean breaking strength of the new fishing line is less than 8 kilograms.

The test statistic for a one-sample t-test is calculated as:

t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))

In this case, the sample mean breaking strength (x̄) is 7.8 kg, the hypothesized mean breaking strength (μ0) is 8 kg, the sample standard deviation (s) is 0.5 kg, and the sample size (n) is 28.

t = (7.8 - 8) / (0.5 / sqrt(28))
t = -0.2 / (0.5 / sqrt(28))
t ≈ -0.2 / (0.0889)
t ≈ -2.25

To determine the p-value, we need to compare the calculated t-value to the t-distribution with n-1 degrees of freedom (df = 28 - 1 = 27), and find the probability of observing a t-value as extreme as the calculated t-value (in the direction of the alternative hypothesis).

Using a t-table or a statistical software, we find that the p-value corresponding to a t-value of -2.25 with 27 degrees of freedom is approximately 0.0158 (or 0.016 when rounded).

The p-value is the probability of obtaining a sample mean breaking strength as extreme as 7.8 kg (or more extreme) assuming that the null hypothesis is true (i.e., the true mean breaking strength is at least 8 kg). Since the p-value (0.016) is less than the significance level (5%), we reject the null hypothesis.

Conclusion: Based on the given data, there is sufficient evidence at the 5% significance level to conclude that the mean breaking strength of the new fishing line is less than 8 kilograms.