The breaking strengths of cables produced by a certain manufacturer have historically had a mean of 1950 pounds and a standard deviation of 85 pounds. The company believes that, due to an improvement in the manufacturing process, the mean breaking strength, , of the cables is now greater than 1950 pounds. To see if this is the case, 90 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1968 pounds. Can we support, at the 0.10 level of significance, the claim that the population mean breaking strength of the newly-manufactured cables is greater than 1950 pounds? Assume that the population standard deviation has not changed.
Perform a one-tailed test. Then complete the parts below.

Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (a)	State the null H0 hypothesis H1 and the alternative hypothesis. (b)	Determine the type of test statistic to use. (c)	Find the value of the test statistic. (Round to three or more decimal places.)

This problem involves conducting a hypothesis test to determine whether the mean breaking strength of the newly-manufactured cables is greater than the historical mean. We'll perform the test step by step.

### Given Information:
- Historical mean breaking strength (\(\mu_0\)): 1950 pounds
- Standard deviation (\(\sigma\)): 85 pounds
- Sample size (\(n\)): 90
- Sample mean breaking strength (\(\bar{x}\)): 1968 pounds
- Significance level (\(\alpha\)): 0.10

### (a) State the Null and Alternative Hypotheses:
- **Null hypothesis (\(H_0\))**: The population mean breaking strength has not increased from 1950 pounds.
  \[
  H_0: \mu \leq 1950
  \]
- **Alternative hypothesis (\(H_1\))**: The population mean breaking strength has increased and is now greater than 1950 pounds.
  \[
  H_1: \mu > 1950
  \]

This is a **right-tailed (one-tailed) test** since we are testing if the mean is greater than 1950.

### (b) Determine the Type of Test Statistic:
Since the population standard deviation is known and the sample size is sufficiently large (\(n = 90\)), we will use a **z-test** (standard normal test).

The test statistic is given by:
\[
z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}
\]
Where:
- \(\bar{x}\) is the sample mean,
- \(\mu_0\) is the population mean under the null hypothesis,
- \(\sigma\) is the population standard deviation,
- \(n\) is the sample size.

### (c) Find the Value of the Test Statistic:
We will now compute the value of the z-statistic using the given values.

\[
z = \frac{1968 - 1950}{\frac{85}{\sqrt{90}}}
\]import math

# Given values
sample_mean = 1968
population_mean = 1950
population_std_dev = 85
sample_size = 90

# Calculate the z-statistic
z_statistic = (sample_mean - population_mean) / (population_std_dev / math.sqrt(sample_size))
z_statisticThe value of the test statistic (z) is approximately **2.009**.

Would you like further details or have any questions about this test? Here are some follow-up inquiries to expand on this topic:

1. What is the critical value for a one-tailed test at the 0.10 significance level?
2. How would you interpret the result based on the z-value?
3. How is the p-value calculated for this test?
4. What assumptions must hold for the z-test to be valid in this scenario?
5. How would the results change if the significance level was set to 0.05?
6. What if the sample size was smaller—how would that impact the test statistic?
7. Can we use a t-test instead of a z-test in a similar scenario, and why?
8. How would a confidence interval approach relate to this hypothesis test?

**Tip**: Always visualize the test statistic against the critical region to reinforce your understanding of hypothesis testing.

Learn how to perform a hypothesis test to determine if the mean breaking strength of newly-manufactured cables is greater than 1950 pounds. This problem involves using the z-test for a one-tailed test with specific data points and significance level.

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