The tires are also tested for failure during a road test where they are subjected to road hazards. A test of 600 tires resulted in the failure of 30 tires.
a) Find a 95% confidence interval on the proportion of failures for the new formula of of rubber.
b) The proportion of failures for the existing line of tires is p=.1 . Can we conclude the new line of tires is better? (use alpha=.05)
To find a confidence interval for the proportion of failures for the new formula of rubber, we can use the formula:
CI = p̂ ± Z * √( p̂(1 - p̂) / n )
where:
- p̂ is the sample proportion of failures (30/600)
- Z is the Z-score corresponding to the desired confidence level (for a 95% confidence level, Z = 1.96)
- n is the sample size (600)
a) Calculating the confidence interval:
p̂ = 30/600 = 0.05
Z = 1.96
n = 600
CI = 0.05 ± 1.96 * √( 0.05(1 - 0.05) / 600 )
Calculating the values within the formula:
p̂(1 - p̂) = 0.05(1 - 0.05) = 0.0475
√( p̂(1 - p̂) / n ) = √( 0.0475 / 600 ) ≈ 0.0109
Substituting the values back into the confidence interval formula:
CI = 0.05 ± 1.96 * 0.0109
Calculating the confidence interval:
CI = (0.05 - 1.96 * 0.0109, 0.05 + 1.96 * 0.0109)
CI ≈ (0.028, 0.072)
Therefore, the 95% confidence interval for the proportion of failures for the new formula of rubber is approximately (0.028, 0.072).
b) To determine if the new line of tires is better than the existing line, we can perform a hypothesis test.
Null hypothesis (H0): The proportion of failures for the new line of tires is equal to the proportion of failures for the existing line (p = 0.1).
Alternative hypothesis (H1): The proportion of failures for the new line of tires is less than the proportion of failures for the existing line (p < 0.1).
Since we are given an alpha value of 0.05, we can reject the null hypothesis if the p-value is less than 0.05.
To calculate the p-value, we can use the Z-test for proportions:
Z = (p̂ - p) / √( (p * (1 - p)) / n )
where:
- p̂ is the sample proportion of failures for the new line of tires (0.05)
- p is the proportion of failures for the existing line of tires (0.1)
- n is the sample size (600)
Calculating the Z-score:
Z = (0.05 - 0.1) / √( (0.1 * (1 - 0.1)) / 600 )
Calculating the values within the formula:
(0.1 * (1 - 0.1)) = 0.09
√( 0.09 / 600 ) ≈ 0.0134
Substituting the values back into the Z-score formula:
Z = -0.05 / 0.0134 ≈ -3.73
Looking up the Z-score in the Z-table, we find that the p-value is less than 0.0001.
Since the p-value (less than 0.0001) is less than the alpha value of 0.05, we can reject the null hypothesis. Therefore, we can conclude that the new line of tires is better than the existing line, as the proportion of failures for the new line is significantly lower.